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Cambridge IGCSE Mathematics 0580 · Core and Extended · Topic 6

Trigonometry

403 free, high-quality past-paper-style questions on Trigonometry for Cambridge IGCSE Maths, each with a full mark scheme. Covers: Bearings; Pythagoras' theorem and right-angled trigonometry; Sine rule, cosine rule, trigonometric graphs and 3D problems. For when you have finished the past papers, or want to practise one topic at a time.Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.2,447 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked

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No calculator

Three villages are labelled A, B and C. The bearing of B from A is 065°. The bearing of C from A is 310°.

Diagram not to scale

(a)

Find the bearing of A from B.

(1)

(b)

Find the bearing of A from C.

(1)

(c)

Find the size of angle BAC.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 065 + 180 = 245°	✓ 1
Part (b)
B1: 310 − 180 = 130°	✓ 1
Part (c)
M1: (360 − 310) + 65, or equivalent method using the north line at A	✓ 1
A1: 115°	✓ 1

Exam tip: For a back bearing, add 180° if the bearing is less than 180°, and subtract 180° if it is more than 180°. The answer must always be between 000° and 360°.

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Calculator

A ship leaves a harbour H and sails 12.5 km on a bearing of 090°. It then sails 8.4 km on a bearing of 180° to a point F.

Diagram not to scale

(a)

Calculate the distance HF.

(2)

(b)

Calculate the bearing of F from H.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: √(12.5² + 8.4²) or √226.81	✓ 1
A1: 15.1 km (15.06…)	✓ 1
Part (b)
M1: tan⁻¹(8.4 ÷ 12.5), or tan⁻¹(12.5 ÷ 8.4), or another correct trig ratio using HF	✓ 1
A1: 33.9° (angle south of east) or 56.1° (angle east of south)	✓ 1
A1: bearing = 90 + 33.9 or 180 − 56.1 = 123.9° (accept 124°)	✓ 1

Common error: The angle 33.9° is measured from the east line, so it is not a bearing. Bearings are measured clockwise from north, so you must add it to 090°. If you use the angle from south (56.1°), subtract it from 180° instead.

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Calculator

A lighthouse L is 6.2 km due north of a harbour H. A boat B is due east of L. The bearing of B from H is 038°.

Diagram not to scale

(a)

Calculate the distance LB.

(2)

(b)

Calculate the distance HB.

(2)

(c)

Write down the bearing of H from B.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: LB = 6.2 × tan 38°	✓ 1
A1: 4.84 km	✓ 1
Part (b)
M1: HB = 6.2 ÷ cos 38°, or Pythagoras using their LB	✓ 1
A1: 7.87 km	✓ 1
Part (c)
B1: 218°	✓ 1

Exam tip: Make a quick sketch with a north line at H. Angle LHB is the bearing 38°, and the right angle is at L because B is due east of L.

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No calculator

This question is about compass directions and maps.

(a)

Write the direction south-west as a three-figure bearing.

(1)

(b)

Town P is north-west of town Q. Write down the bearing of Q from P.

(1)

(c)

A map has a scale of 1 : 25 000. On the map, the distance from P to Q is 18 cm. Find the actual distance from P to Q in kilometres.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 225°	✓ 1
Part (b)
B1: 135°	✓ 1
Part (c)
M1: 18 × 25 000 = 450 000 cm	✓ 1
A1: 4.5 km	✓ 1

Common error: In the second part, Q is south-east of P, not north-west. Read 'the bearing of Q from P' as 'stand at P and face Q'.

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Calculator

A hiker starts at S and walks 3.6 km due west. She then walks 5.1 km due north to a point E.

Diagram not to scale

(a)

Calculate the direct distance SE.

(2)

(b)

Calculate the bearing of S from E.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: \(\sqrt{3.6^2 + 5.1^2}\) (= \(\sqrt{38.97}\))	✓ 1
A1: 6.24 km (accept answers rounding to 6.24)	✓ 1
Part (b)
M1: a correct trigonometric method for an angle at E, e.g. tan⁻¹(3.6 ÷ 5.1), or tan⁻¹(5.1 ÷ 3.6), or sin⁻¹(3.6 ÷ 6.24)	✓ 1
A1: correct angle, e.g. 35.2° (from south towards east) or 54.8° (from east towards south)	✓ 1
A1: bearing = 180 − 35.2 or 90 + 54.8 = 144.8° (accept answers rounding to 145°)	✓ 1

Common error: The question asks for the bearing of S from E, so draw the north line at E. S lies south-east of E, so the bearing must be between 090° and 180°. Do not give the bearing of E from S instead.

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Calculator

A ladder of length 6.5 m leans against a vertical wall. The foot of the ladder is on horizontal ground, 1.8 m from the base of the wall.

Diagram not to scale

(a)

Calculate how far up the wall the ladder reaches.

(3)

(b)

Calculate the angle the ladder makes with the ground.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 6.5² − 1.8² (= 39.01)	✓ 1
M1: √(6.5² − 1.8²)	✓ 1
A1: 6.25 m (6.2458…)	✓ 1
Part (b)
M1: cos θ = 1.8 ÷ 6.5, or sin θ = 6.2458… ÷ 6.5, or tan θ = 6.2458… ÷ 1.8, using the unrounded height from the previous part	✓ 1
A1: 73.9°	✓ 1

Common error: The ladder is the hypotenuse, so subtract the squares. Adding them gives 6.74 m, which is longer than the ladder itself. If you use the height to find the angle, keep its unrounded value: sin θ = 6.25 ÷ 6.5 gives 74.0°, not 73.9°.

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No calculator

A right-angled triangle has shorter sides of 9 cm and 12 cm. The angle opposite the 9 cm side is x.

Diagram not to scale

(a)

Find the length of the hypotenuse.

(2)

(b)

Write cos x as a fraction in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 9² + 12² = 81 + 144 = 225	✓ 1
A1: 15 cm	✓ 1
Part (b)
B1: \(\frac{12}{15}\) seen	✓ 1
B1: \(\frac{4}{5}\)	✓ 1

Exam tip: Label the sides opposite, adjacent and hypotenuse relative to angle x before you choose a ratio. The 12 cm side is adjacent to x.

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Calculator

Triangle PQR has a right angle at Q.
PR = 14.2 cm and angle QPR = 38°.

Diagram not to scale

(a)

Calculate QR.

(2)

(b)

Calculate PQ.

(2)

(c)

Calculate the area of triangle PQR.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 14.2 × sin 38°	✓ 1
A1: 8.74 cm (8.742…)	✓ 1
Part (b)
M1: 14.2 × cos 38°	✓ 1
A1: 11.2 cm (11.189…)	✓ 1
Part (c)
M1: ½ × their QR × their PQ	✓ 1
A1: 48.9 cm²	✓ 1

Exam tip: Keep the unrounded values from parts (a) and (b) in your calculator for part (c). This avoids rounding errors in the final answer.

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No calculator

(a)

A triangle has sides 7 cm, 24 cm and 25 cm. Show that the triangle is right-angled.

(2)

(b)

A different right-angled triangle has a hypotenuse of 13 cm and one side of 5 cm. Find the length of the third side.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 7² + 24² = 49 + 576 = 625	✓ 1
A1: 25² = 625, so by Pythagoras the triangle is right-angled	✓ 1
Part (b)
M1: 13² − 5² = 169 − 25 = 144	✓ 1
A1: 12 cm	✓ 1

Exam tip: In a 'show that' question you must write a conclusion. State that the two values are equal, so the triangle is right-angled.

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Calculator

A is the point (−3, 5) and B is the point (6, −1).

(a)

Calculate the length of AB.

(3)

(b)

Calculate the acute angle that the line AB makes with the x-axis.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: horizontal change 9 and vertical change 6	✓ 1
M1: √(9² + 6²) = √117	✓ 1
A1: 10.8 units (10.81…)	✓ 1
Part (b)
M1: tan θ = 6 ÷ 9	✓ 1
A1: 33.7°	✓ 1

Common error: Subtracting −3 from 6 gives 9, not 3. Take extra care with negative coordinates when you find the differences.

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Calculator

A window cleaner leans a ladder of length 4.5 m against a vertical wall. The foot of the ladder is on level ground, 1.2 m from the wall.

Diagram not to scale

(a)

Calculate how high up the wall the ladder reaches. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate the angle between the ladder and the ground.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 correct Pythagoras statement, e.g. h² + 1.2² = 4.5²	✓ 1
M1 h = √(4.5² − 1.2²) = √18.81	✓ 1
A1 4.34 m	✓ 1
Part (b)
M1 cos θ = 1.2 ÷ 4.5 (or sin θ = 4.337 ÷ 4.5 or tan θ = 4.337 ÷ 1.2, ECF from (a))	✓ 1
A1 74.5°	✓ 1

Common error: When you rearrange Pythagoras' theorem to find a shorter side, subtract the squares; adding them gives the hypotenuse instead.

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Calculator

A straight wheelchair ramp rises through a vertical height of 0.45 m. The ramp makes an angle of 4.8° with the horizontal ground.

Diagram not to scale

(a)

Calculate the length of the ramp. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate the horizontal distance covered by the ramp. Use unrounded values and give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 identifies the ramp as the hypotenuse, with 0.45 m opposite the 4.8° angle	✓ 1
M1 sin 4.8° = 0.45 ÷ L (or L = 0.45 ÷ sin 4.8°)	✓ 1
A1 5.38 m	✓ 1
Part (b)
M1 tan 4.8° = 0.45 ÷ d (or d = L cos 4.8°, or Pythagoras with 0.45 and L; ECF from (a))	✓ 1
A1 5.36 m	✓ 1

Exam tip: Label the sides opposite, adjacent and hypotenuse relative to the given angle before choosing sin, cos or tan.

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Calculator

A student stands 85 m from the base of a wind turbine on level ground. Her eyes are 1.6 m above the ground. She measures the angle of elevation of the top of the turbine blade, at its highest point, as 52°.

Diagram not to scale

(a)

Calculate the height of the blade tip above the level of her eyes. Give your answer correct to 1 decimal place.

(2)

(b)

Calculate the height of the blade tip above the ground. Give your answer correct to 1 decimal place.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan 52° = height ÷ 85	✓ 1
A1 108.8 m	✓ 1
Part (b)
B1 110.4 m (ECF from (a): add 1.6 m)	✓ 1

Exam tip: Angles of elevation are measured from the horizontal at the observer's eye, so remember to add the observer's eye height at the end.

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Calculator

A rectangular television screen has a diagonal of 140 cm and a width of 122 cm.

Diagram not to scale

(a)

Calculate the height of the screen. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the angle between the diagonal and the width of the screen. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 h² = 140² − 122² (= 4716)	✓ 1
A1 68.7 cm	✓ 1
Part (b)
M1 cos θ = 122 ÷ 140 (or sin θ = 68.7 ÷ 140 or tan θ = 68.7 ÷ 122)	✓ 1
A1 29.4°	✓ 1

Common error: The diagonal is the longest side (the hypotenuse), so it must be the 140 in the squared equation with the other side subtracted.

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Calculator

A water pipe is laid down a slope. It covers a horizontal distance of 12 m and drops a vertical height of 5 m.

Diagram not to scale

(a)

Calculate the length of the pipe.

(2)

(b)

The pipe costs €8.50 per metre. Calculate the total cost of the pipe.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(12² + 5²)	✓ 1
A1 13 m	✓ 1
Part (b)
M1 13 × 8.50 (ECF from (a))	✓ 1
A1 €110.50	✓ 1

Exam tip: Use the unrounded length from (a) in later calculations; here it is exactly 13 m.

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Calculator

A rectangular game screen is 30 cm wide and its diagonal is 34 cm.

Diagram not to scale

(a)

Calculate the height of the screen.

(3)

(b)

Calculate the area of the screen.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 height² + 30² = 34²	✓ 1
M1 √(34² − 30²) = √256	✓ 1
A1 16 cm	✓ 1
Part (b)
B1 30 × 16 = 480 cm² (ECF from (a))	✓ 1

Common error: When the longest side (hypotenuse) is known, subtract the squares; adding them gives a length longer than the hypotenuse.

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Calculator

A pumping station P is 6 km from a water tower W on a bearing of 070° from W.

Diagram not to scale

(a)

Write down the bearing of W from P.

(2)

(b)

Calculate how far north of W the pumping station is, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 070° + 180°	✓ 1
A1 250°	✓ 1
Part (b)
M1 6 × cos 70°	✓ 1
A1 2.05 km	✓ 1

Exam tip: The back-bearing is the original bearing plus or minus 180°, giving an answer between 0° and 360°.

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Calculator

A 4.5 m ladder leans against a vertical wall beside a water tank. The ladder makes an angle of 65° with the horizontal ground.

Diagram not to scale

(a)

Calculate the height the ladder reaches up the wall, to 3 significant figures.

(2)

(b)

Calculate the distance from the foot of the ladder to the wall, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 4.5 × sin 65°	✓ 1
A1 4.08 m	✓ 1
Part (b)
M1 4.5 × cos 65°	✓ 1
A1 1.90 m	✓ 1

Common error: Opposite the angle is the side away from the angle: with the angle at the ground, the wall height is opposite, so use sine.

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Calculator

A light aircraft leaves a grass airstrip and flies 240 km due east. It then flies 70 km due north and lands. Both legs are straight and the ground is treated as flat.

Diagram not to scale

(a)

Calculate the distance of the landing point from the airstrip.

(2)

(b)

Calculate the bearing of the landing point from the airstrip. Give your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses Pythagoras: 240² + 70² = 62 500	✓ 1
Distance = 250 km	✓ 1
Part (b)
Angle measured from north: tan θ = 240/70	✓ 1
θ = 73.7°, so bearing 073.7°	✓ 1

Common error: Bearings are measured clockwise from north, so decide which side of the right-angled triangle is opposite the angle before choosing tan.

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Calculator

A conveyor belt carries sacks of grain from the ground up to the loading hatch of a barn. The belt is straight, 12 m long and rises at a constant angle of 25° to the horizontal.

Diagram not to scale

(a)

Calculate the height of the loading hatch above the ground. Give your answer to 3 significant figures.

(2)

(b)

Calculate the horizontal distance between the bottom of the belt and the barn wall directly below the hatch. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses sin 25° = height/12	✓ 1
Height = 5.07 m	✓ 1
Part (b)
Uses cos 25° = horizontal/12 (or Pythagoras with their height)	✓ 1
Horizontal distance = 10.9 m	✓ 1

Exam tip: Write down which ratio links the known side, the unknown side and the angle (SOH, CAH or TOA) before you substitute numbers.

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Calculator

A drone flies in a straight line from point X to a farm at point Y, a distance of 450 m on a bearing of 320°.

Diagram not to scale

(a)

Write down the bearing of X from Y.

(1)

(b)

Calculate how far west of X the farm is. Give your answer to 3 significant figures.

(2)

(c)

Calculate how far north of X the farm is. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
320° − 180° = 140°	✓ 1
Part (b)
Angle between flight path and north is 40° (360° − 320°); west distance = 450 sin 40°	✓ 1
West distance = 289 m	✓ 1
Part (c)
North distance = 450 cos 40°	✓ 1
North distance = 345 m	✓ 1

Common error: Bearing 320° is 40° west of north, not 20° or 50°; subtract from 360° first.

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Calculator

A roof frame for a hay barn is a triangle with sides of length 7.5 m, 10 m and 12.5 m.

Diagram not to scale

(a)

Show that the triangle is right-angled.

(2)

(b)

Calculate the smallest angle of the triangle. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
7.5² + 10² = 156.25	✓ 1
12.5² = 156.25, so by the converse of Pythagoras' theorem the triangle is right-angled	✓ 1
Part (b)
Smallest angle is opposite the 7.5 m side: sin θ = 7.5/12.5	✓ 1
θ = 36.9°	✓ 1

Exam tip: The hypotenuse is the longest side and sits opposite the right angle; test the two shorter sides squared against it.

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No calculator

A supply ship sails 15 km due west from a port and then 8 km due south to reach an oil platform.

Diagram not to scale

(a)

Calculate the straight-line distance from the port to the platform.

(2)

(b)

The ship returns directly to the port at a constant 12 km/h. Calculate the time taken, in minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses Pythagoras: 15² + 8² = 289	✓ 1
Distance = 17 km	✓ 1
Part (b)
Time = 17/12 hours	✓ 1
17/12 × 60 = 85 minutes	✓ 1

Exam tip: Learn the common Pythagorean triples (such as 8, 15, 17) so you can spot exact answers when no calculator is allowed.

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Calculator

A volcanologist stands on level ground 3.6 km horizontally from the point directly below the top of an ash plume. She measures the angle of elevation of the top of the plume as 38°. Ignore her height.

Diagram not to scale

(a)

Calculate the height of the top of the plume above the ground. Give your answer to 3 significant figures.

(2)

(b)

Calculate the distance from the volcanologist to the top of the plume. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses tan 38° = height/3.6	✓ 1
Height = 2.81 km	✓ 1
Part (b)
Uses cos 38° = 3.6/distance (or Pythagoras with their height)	✓ 1
Distance = 4.57 km	✓ 1

Exam tip: Check a distance answer: the hypotenuse must be longer than both of the other sides.

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Calculator

A gangway 6.5 m long joins a quay to the deck of a ferry. The deck is 2.5 m above the level of the quay and the gangway is straight.

Diagram not to scale

(a)

Calculate the horizontal distance from the foot of the gangway to the point on the quay directly below the deck edge it reaches.

(2)

(b)

Calculate the angle the gangway makes with the horizontal. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Pythagoras: 6.5² − 2.5² = 36	✓ 1
Horizontal distance = 6 m	✓ 1
Part (b)
sin θ = 2.5/6.5	✓ 1
θ = 22.6°	✓ 1

Common error: When the hypotenuse is given, subtract the square of the known shorter side from the square of the hypotenuse; adding them would give a length longer than the gangway.

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Calculator

An archaeology team marks out a rectangular trench. The width of the trench is 3.6 m and a taut string stretched from one corner to the opposite corner is 8.5 m long.

Diagram not to scale

(a)

Calculate the length of the trench.

(2)

(b)

Calculate the perimeter of the trench.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of Pythagoras, length² = 8.5² − 3.6²	✓ 1
A1 7.70 m (7.7)	✓ 1
Part (b)
M1 2 × (7.7 + 3.6) or 2 × 7.7 + 2 × 3.6, ECF from (a)	✓ 1
A1 22.6 m	✓ 1

Common error: When the hypotenuse is given, subtract the squares. Adding them gives a length longer than the diagonal, which is impossible.

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Calculator

A cycle ramp at a skatepark is a straight slope 12 m long. The top of the ramp is 1.5 m higher than the bottom.

Diagram not to scale

(a)

Calculate the angle the ramp makes with the horizontal. Give your answer correct to 1 decimal place.

(2)

(b)

Calculate the horizontal distance between the bottom and the top of the ramp. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sin x = 1.5 ÷ 12	✓ 1
A1 7.2°	✓ 1
Part (b)
M1 use of Pythagoras, 12² − 1.5²	✓ 1
A1 11.9 m	✓ 1

Exam tip: Sketch the right-angled triangle and label the hypotenuse first. Then choose sin, cos or tan from the two sides you know.

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Calculator

A cycle route starts at point S and runs on a bearing of 048°. At a junction the route turns right (clockwise) through 75°. Later it turns left (anticlockwise) through 140°.

Diagram not to scale

(a)

Find the bearing of the route after the first turn.

(1)

(b)

Find the bearing of the route after the second turn.

(2)

(c)

A cyclist at the junction of the first turn wants to ride straight back to S along the first stage of the route. Find the bearing she must follow.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 123°	✓ 1
Part (b)
M1 123° − 140° = −17°	✓ 1
A1 343°	✓ 1
Part (c)
M1 048° + 180°	✓ 1
A1 228°	✓ 1

Exam tip: A bearing must always be written with three figures, for example 048°, and lies between 000° and 360°. Add or subtract 360° if your answer falls outside that range.

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Calculator

A ladder rests against a vertical wall of an excavation. The foot of the ladder is on level ground 1.2 m from the wall. The top of the ladder touches the wall at a height of 3.5 m above the ground.

Diagram not to scale

(a)

Calculate the length of the ladder.

(2)

(b)

Calculate the angle the ladder makes with the ground. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1.2² + 3.5² = 13.69	✓ 1
A1 3.70 m (3.7)	✓ 1
Part (b)
M1 tan x = 3.5 ÷ 1.2 (or cos x = 1.2 ÷ 3.7)	✓ 1
A1 71.1°	✓ 1

Exam tip: Choose the trig ratio that uses the two sides you know. Here the side next to the angle and the side opposite it are given, so use tan.

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Calculator

A swimming pool is a rectangle 25 m long and 10 m wide. A swimmer crosses the pool along the straight diagonal from one corner to the opposite corner.

Diagram not to scale

(a)

Calculate the distance the swimmer travels. Give your answer correct to 3 significant figures.

(2)

(b)

The swimmer travels at a constant speed of 1.25 m/s. Calculate the time taken, using the unrounded distance from (a). Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 25² + 10² = 725	✓ 1
A1 26.9 m	✓ 1
Part (b)
M1 time = distance ÷ speed = √725 ÷ 1.25	✓ 1
A1 21.5 s	✓ 1

Common error: Add the squares of the two shorter sides to find the diagonal. Do not add the sides themselves, which would give 35 m.

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Calculator

A control hut H stands on a wind farm. Turbine T is on a bearing of 075° from H. The bearing of H from a second turbine U is 318°.

Diagram not to scale

(a)

Find the bearing of H from T.

(2)

(b)

Find the bearing of U from H.

(1)

(c)

Find angle THU.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 075° + 180°	✓ 1
A1 255°	✓ 1
Part (b)
B1 318° − 180° = 138°	✓ 1
Part (c)
M1 138° − 075°	✓ 1
A1 63°	✓ 1

Exam tip: The bearing from B to A and the bearing from A to B differ by 180°. Add 180° if the bearing is below 180°, and subtract 180° if it is above.

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Calculator

A library entrance has a straight wheelchair ramp. The ramp rises 0.45 m and covers a horizontal distance of 5.4 m.

Diagram not to scale

(a)

Calculate the length of the ramp. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the angle of the ramp to the horizontal. Give your answer correct to 1 decimal place.

(2)

(c)

A guideline says the angle of a ramp should be less than 5°. State whether this ramp meets the guideline.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5.4² + 0.45²	✓ 1
A1 5.42 m	✓ 1
Part (b)
M1 tan x = 0.45 ÷ 5.4	✓ 1
A1 4.8°	✓ 1
Part (c)
B1 Yes, 4.8° is less than 5°	✓ 1

Exam tip: In a final comment part, quote your value and compare it with the limit. A bare yes or no does not earn the mark.

+ Add to test✓ In your test33IGMC3-6-10r7-10Pythagoras' theorem and right-angled trigonometry Paper CEasy4 marks

Calculator

A library hangs an isosceles triangular banner. The base is 1.8 m and each of the two equal sides is 2.5 m.

Diagram not to scale

(a)

Calculate the perpendicular height of the banner. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the area of the banner. Use the unrounded height and give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 the height splits the base into two halves of 0.9 m, so height² = 2.5² − 0.9²	✓ 1
A1 2.33 m	✓ 1
Part (b)
M1 area = ½ × 1.8 × height	✓ 1
A1 2.10 m²	✓ 1

Common error: The height of an isosceles triangle meets the base at its midpoint. Using the full base of 1.8 m instead of half in the Pythagoras step gives the wrong height.

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Calculator

A climbing path rises at a constant slope. A mountaineer walks 150 m along the path and gains 90 m in height.

Diagram not to scale

(a)

Calculate the horizontal distance covered.

(2)

(b)

Calculate the angle the path makes with the horizontal. Give your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(150² − 90²)	✓ 1
A1 120 m	✓ 1
Part (b)
M1 sin⁻¹(90/150)	✓ 1
A1 36.9°	✓ 1

Common error: Using the horizontal distance instead of the path length as the hypotenuse.

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Calculator

A telecoms mast stands on level ground. From a point 45 m from its base, the angle of elevation of the top is 36°.

Diagram not to scale

(a)

Calculate the height of the mast. Give your answer to 3 significant figures.

(2)

(b)

A support wire runs from the top of the mast to the point 45 m from the base. Calculate its length, using unrounded values. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 45 × tan 36°	✓ 1
A1 32.7 m	✓ 1
Part (b)
M1 45 ÷ cos 36° (or √(45² + height²))	✓ 1
A1 55.6 m	✓ 1

Exam tip: Choose the trigonometric ratio that uses the data you were given, not a value you calculated earlier.

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Calculator

A mountain hut H is 3 km from a camp C, on a bearing of 065° from C.

Diagram not to scale

(a)

Write down the bearing of C from H.

(1)

(b)

Calculate how far east of C the hut is. Give your answer to 3 significant figures.

(2)

(c)

Calculate how far north of C the hut is. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 245°	✓ 1
Part (b)
M1 3 × sin 65°	✓ 1
A1 2.72 km	✓ 1
Part (c)
M1 3 × cos 65°	✓ 1
A1 1.27 km	✓ 1

Common error: Mixing up sine and cosine; with a bearing measured from north, the northward component uses cosine.

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Calculator

A conical coffee filter has a slant height of 13 cm and a circular opening of radius 5 cm.

Cross-section, not to scale

(a)

Calculate the vertical height of the cone.

(2)

(b)

Calculate the angle between the slant edge and the vertical axis of the cone. Give your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(13² − 5²)	✓ 1
A1 12 cm	✓ 1
Part (b)
M1 tan⁻¹(5/12)	✓ 1
A1 22.6°	✓ 1

Exam tip: The slant height is the hypotenuse of the right-angled triangle formed by the height and the radius.

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Calculator

A tracking mast on a space probe's landing pad is 6.0 m tall and stands vertically on flat ground. A straight support cable runs from the top of the mast to a point on the ground 4.5 m from the base of the mast.

Diagram not to scale

(a)

Calculate the length of the cable.

(2)

(b)

Calculate the angle between the cable and the ground. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of Pythagoras: length² = 6.0² + 4.5²	✓ 1
A1 7.5 m	✓ 1
Part (b)
M1 tan x = 6.0 ÷ 4.5	✓ 1
A1 53.1°	✓ 1

Exam tip: When both shorter sides are known, tan gives the angle directly without relying on a rounded cable length.

+ Add to test✓ In your test39IGMC3-6-09r9-02Pythagoras' theorem and right-angled trigonometry Paper CEasy4 marks

Calculator

A conveyor belt carries green coffee beans up to a roasting drum. The belt is a straight line 3.2 m long and rises at 18° to the horizontal.

Diagram not to scale

(a)

Calculate the vertical height gained by the beans. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the horizontal distance covered by the belt. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3.2 × sin 18°	✓ 1
A1 0.989 m	✓ 1
Part (b)
M1 3.2 × cos 18°	✓ 1
A1 3.04 m	✓ 1

Common error: Mixing up sin and cos for the height and the horizontal distance. The side opposite the angle is the height.

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Calculator

A rover leaves its lander and drives 12 km due east, then 5 km due north to reach a crater.

Diagram not to scale

(a)

Calculate the straight-line distance from the lander to the crater.

(2)

(b)

Find the bearing of the crater from the lander. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12² + 5² (= 169)	✓ 1
A1 13 km	✓ 1
Part (b)
M1 angle north of east: tan x = 5 ÷ 12	✓ 1
A1 x = 22.6°	✓ 1
A1 bearing 067.4° (90 − 22.6)	✓ 1

Common error: Giving 22.6° as the bearing. A bearing is measured clockwise from north, so subtract from 90°.

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Calculator

A coffee freighter sails in a straight line from port P to port Q. The bearing of Q from P is 134°. The distance PQ is 90 km.

Diagram not to scale

(a)

Find the bearing of P from Q.

(2)

(b)

Calculate how far east of P the port Q is. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 134 + 180 (or co-interior angles 46°)	✓ 1
A1 314°	✓ 1
Part (b)
M1 90 × sin 46° (or 90 × sin 134°)	✓ 1
A1 64.7 km	✓ 1

Exam tip: For a back bearing, add 180° to a bearing under 180° and subtract 180° from a bearing over 180°.

+ Add to test✓ In your test42IGMC3-6-09r10-01Pythagoras' theorem and right-angled trigonometry Paper CEasy4 marks

Calculator

A conveyor belt in a grain store is 8.5 m long and rises at a constant angle of 32° to the horizontal. Give answers to 3 significant figures.

Diagram not to scale

(a)

Calculate the vertical height the belt lifts the grain.

(2)

(b)

Calculate the horizontal distance covered by the belt.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sin 32° = h/8.5	✓ 1
A1 h = 4.50 m	✓ 1
Part (b)
M1 cos 32° = d/8.5	✓ 1
A1 d = 7.21 m	✓ 1

Common error: Check that the calculator is in degree mode before using sin, cos or tan.

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Calculator

An aircraft in a steady descent covers a horizontal distance of 12 km while losing 1.8 km of altitude. Its path is a straight line.

Diagram not to scale

(a)

Calculate the distance flown along the descent path. Give your answer to 3 significant figures.

(2)

(b)

Calculate the angle of descent below the horizontal. Give your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 distance² = 12² + 1.8²	✓ 1
A1 12.1 km	✓ 1
Part (b)
M1 tan θ = 1.8/12	✓ 1
A1 8.5°	✓ 1

Exam tip: Use the two sides you were given rather than a rounded answer from an earlier part.

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No calculator

On a farm map, the barn B is on a bearing of 072° from the farmhouse F. A hay store H is on a bearing of 215° from F.

Diagram not to scale

(a)

Find the bearing of F from B.

(2)

(b)

Find the bearing of F from H.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 180 + 72 (or 72 + 180)	✓ 1
A1 252°	✓ 1
Part (b)
M1 215 − 180	✓ 1
A1 35°	✓ 1

Common error: To find a back bearing add 180° if the bearing is less than 180°, and subtract 180° if it is more.

+ Add to test✓ In your test45IGMC3-6-09r10-04Pythagoras' theorem and right-angled trigonometry Paper CEasy4 marks

No calculator

A rectangular stage measures 12 m by 5 m. A performer walks in a straight line from one corner to the opposite corner.

Diagram not to scale

(a)

Find the length of the diagonal of the stage.

(2)

(b)

The performer walks at 1.3 m/s. Find the time taken to walk the diagonal.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12² + 5² = 144 + 25 = 169	✓ 1
A1 13 m	✓ 1
Part (b)
M1 time = 13 ÷ 1.3	✓ 1
A1 10 s	✓ 1

Exam tip: Learn the triples 3, 4, 5 and 5, 12, 13 so that no-calculator Pythagoras questions are quick.

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Calculator

A theatre builds a straight ramp onto its stage. The ramp rises 0.90 m vertically over a horizontal distance of 4.00 m.

Diagram not to scale

(a)

Calculate the length of the sloping surface of the ramp.

(2)

(b)

Calculate the angle the ramp makes with the horizontal. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 length² = 0.90² + 4.00² (or sqrt(0.90² + 4.00²))	✓ 1
A1 4.10 m	✓ 1
Part (b)
M1 tan⁻¹(0.90 ÷ 4.00) (or equivalent using the sloping length)	✓ 1
A1 12.7°	✓ 1

Exam tip: Use the two sides you were given rather than a rounded answer from an earlier part.

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Calculator

A spotlight is fixed to a lighting bar 6.5 m above a stage floor. The beam is aimed at a spot on the floor that is 4.2 m horizontally from the point directly below the spotlight.

Diagram not to scale

(a)

Calculate the length of the beam from the spotlight to the spot. Give your answer to 3 significant figures.

(2)

(b)

Calculate the angle of depression of the spot as seen from the spotlight. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sqrt(6.5² + 4.2²)	✓ 1
A1 7.74 m	✓ 1
Part (b)
M1 tan⁻¹(6.5 ÷ 4.2)	✓ 1
A1 57.1°	✓ 1

Exam tip: The angle of depression is measured from the horizontal, down to the line of sight.

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Calculator

The lamp of a lighthouse is 36 m above sea level. A boat is on the sea, level with the base of the lighthouse. From the lamp, the angle of depression of the boat is 4.5°.

Diagram not to scale

(a)

Calculate the horizontal distance from the boat to the base of the lighthouse. Give your answer to the nearest metre.

(2)

(b)

The boat sails towards the lighthouse until it is 300 m from the base. Calculate the new angle of depression of the boat from the lamp. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 36 ÷ tan 4.5°	✓ 1
A1 457 m	✓ 1
Part (b)
M1 tan⁻¹(36 ÷ 300)	✓ 1
A1 6.8°	✓ 1

Common error: The angle of depression from the lamp equals the angle of elevation of the lamp from the boat, not the angle at the lamp inside the triangle.

+ Add to test✓ In your test49IGMC3-6-09r11-09Pythagoras' theorem and right-angled trigonometry Paper CEasy6 marks

Calculator

An ash plume rises from a vent. Its top is 3.2 km above the level of an observer's eye. The observer is 7.5 km horizontally from the point directly below the top of the plume.

Diagram not to scale

(a)

Calculate the angle of elevation of the top of the plume from the observer. Give your answer correct to 1 decimal place.

(2)

(b)

Calculate the straight-line distance from the observer to the top of the plume. Give your answer to 3 significant figures.

(2)

(c)

The plume keeps rising until its top has an angle of elevation of 30° from the observer. Calculate its height above the observer's eye level. Give your answer correct to 2 decimal places.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan⁻¹(3.2 ÷ 7.5)	✓ 1
A1 23.1°	✓ 1
Part (b)
M1 sqrt(7.5² + 3.2²)	✓ 1
A1 8.15 km	✓ 1
Part (c)
M1 7.5 × tan 30°	✓ 1
A1 4.33 km	✓ 1

Exam tip: Label the opposite, adjacent and hypotenuse relative to the angle before choosing sin, cos or tan.

+ Add to test✓ In your test50IGMC3-6-08r14-01Pythagoras' theorem and right-angled trigonometry Paper CEasy4 marks

Calculator

A water pipeline climbs a hill. Over a horizontal distance of 840 m the pipe rises vertically by 135 m. Give answers to 3 significant figures, and angles to 1 decimal place.

Diagram not to scale

(a)

Calculate the length of the pipe.

(2)

(b)

Calculate the angle the pipe makes with the horizontal.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sqrt(840² + 135²)	✓ 1
A1 851 m	✓ 1
Part (b)
M1 tan θ = 135/840 (or equivalent using the length from (a))	✓ 1
A1 9.1°	✓ 1

Exam tip: Check that your angle is small for a gentle slope: the rise is much smaller than the horizontal distance.

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Calculator

A jump ramp in a video game is a straight slope of length 18.5 m, inclined at 24° to the horizontal ground. Give answers to 3 significant figures.

Diagram not to scale

(a)

Calculate the height of the top of the ramp above the ground.

(2)

(b)

Calculate the horizontal distance covered by the ramp.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 18.5 sin 24°	✓ 1
A1 7.52 m	✓ 1
Part (b)
M1 18.5 cos 24°	✓ 1
A1 16.9 m	✓ 1

Common error: Make sure the calculator is in degree mode before using sin, cos or tan.

+ Add to test✓ In your test52IGMC3-6-08r14-04Pythagoras' theorem and right-angled trigonometry Paper CEasy4 marks

No calculator

Water is pumped from a tank at A. Pipes run 24 m due east from A to B and then 7 m due north from B to C. A new straight pipe is laid directly from A to C.

Diagram not to scale

(a)

Calculate the length AC.

(2)

(b)

Write down the exact value of cos of angle BAC.

(1)

(c)

How much shorter is the direct pipe than the route through B?

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sqrt(24² + 7²) = sqrt(625)	✓ 1
A1 25 m	✓ 1
Part (b)
B1 24/25	✓ 1
Part (c)
B1 (24 + 7) − 25 = 6 m	✓ 1

Exam tip: Without a calculator, look for whole-number triples such as 7, 24, 25 when working with Pythagoras.

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Calculator

A pharmacy stocks syrup in a cylindrical bottle with an internal diameter of 6.0 cm and an internal height of 14.0 cm. A thin straight rod rests in the bottle with one end at the bottom edge and the other touching the opposite point on the top rim, in a vertical plane through the axis of the bottle.

Diagram not to scale

(a)

Calculate the length of the rod, to 3 significant figures.

(2)

(b)

Calculate the angle between the rod and the base of the bottle, to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sqrt(6.0² + 14.0²)	✓ 1
A1 15.2 cm	✓ 1
Part (b)
M1 tan x = 14.0/6.0	✓ 1
A1 66.8°	✓ 1

Exam tip: Draw the rectangle formed by the vertical section through the axis; the rod is its diagonal.

+ Add to test✓ In your test54IGMC3-6-08r18-01Pythagoras' theorem and right-angled trigonometry Paper CEasy4 marks

Calculator

A bridge tower stands vertically on a level deck. The top of the tower is joined by a straight cable to an anchor point on the deck, 36 m from the foot of the tower. The tower is 15 m high.

Diagram not to scale

(a)

Calculate the length of the cable.

(2)

(b)

Calculate the angle between the cable and the deck. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cable² = 15² + 36²	✓ 1
A1 39 m	✓ 1
Part (b)
M1 tan x = 15 / 36	✓ 1
A1 22.6°	✓ 1

Common error: Check that the side opposite the angle is the 15 m tower, not the 36 m deck length.

+ Add to test✓ In your test55IGMC3-6-08r18-02Pythagoras' theorem and right-angled trigonometry Paper CEasy4 marks

No calculator

A rectangular solar panel on a space probe is 1.5 m wide and 3.6 m long.

Diagram not to scale

(a)

Find the length of a diagonal of the panel.

(2)

(b)

A second rectangular panel has a diagonal of 6.5 m and a width of 2.5 m. Find its length.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1.5² + 3.6² = 2.25 + 12.96 = 15.21	✓ 1
A1 3.9 m	✓ 1
Part (b)
M1 6.5² − 2.5² = 42.25 − 6.25 = 36	✓ 1
A1 6 m	✓ 1

Exam tip: When finding a shorter side, subtract the squares; add them only when finding the longest side.

+ Add to test✓ In your test56IGMC3-6-08r18-05Bearings Paper CEasy5 marks

No calculator

A coffee warehouse W has a farm F on a bearing of 072° from it and a mill M on a bearing of 200° from it. The distance WF is equal to the distance WM.

Diagram not to scale

(a)

Find the bearing of W from F.

(2)

(b)

Find angle FWM.

(1)

(c)

Find angle WFM.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 072° + 180°	✓ 1
A1 252°	✓ 1
Part (b)
B1 200° − 72° = 128°	✓ 1
Part (c)
M1 (180° − 128°) / 2	✓ 1
A1 26°	✓ 1

Common error: For a back bearing add 180° when the first bearing is less than 180°, and subtract 180° when it is greater.

+ Add to test✓ In your test57IGMC3-6-08r18-08Pythagoras' theorem and right-angled trigonometry Paper CEasy4 marks

Calculator

A lander carries a straight ramp of length 2.6 m for a rover. The top of the ramp is 0.45 m above level ground.

Diagram not to scale

(a)

Calculate the angle between the ramp and the ground. Give your answer correct to 1 decimal place.

(2)

(b)

Calculate the horizontal distance from the foot of the ramp to the point directly below its top. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sin x = 0.45 / 2.6	✓ 1
A1 10.0°	✓ 1
Part (b)
M1 distance² = 2.6² − 0.45² = 6.5575	✓ 1
A1 2.56 m	✓ 1

Exam tip: The ramp is the longest side, so its length is the hypotenuse.

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Calculator

A vertical wind turbine mast stands on level ground. A straight support cable runs from a point on the mast 24 m above the ground to a point on the ground 7 m from the foot of the mast.

Diagram not to scale

(a)

Calculate the length of the cable.

(2)

(b)

Calculate the angle that the cable makes with the ground. Give your answer correct to 1 decimal place.

(2)

(c)

The cable costs £4.80 per metre. Three identical cables are used to support the mast. Calculate the total cost of the cables.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{24^2+7^2}\) or \(\sqrt{625}\)	✓ 1
A1 25 m	✓ 1
Part (b)
M1 \(\tan^{-1}\left(\frac{24}{7}\right)\) (or \(\sin^{-1}\frac{24}{25}\) or \(\cos^{-1}\frac{7}{25}\))	✓ 1
A1 73.7°	✓ 1
Part (c)
M1 \(3\times25\times4.80\) (ECF from (a))	✓ 1
A1 £360	✓ 1

Common error: Choosing sin, cos or tan before labelling the opposite, adjacent and hypotenuse sides leads to the wrong ratio; label the triangle first.

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Calculator

A mobile vaccination van leaves a depot D and drives 15 km due east to a village A. It then drives 8 km due north to a village B.

Diagram not to scale

(a)

Calculate the distance DB.

(2)

(b)

Calculate the bearing of B from D. Give your answer correct to 1 decimal place.

(2)

(c)

Find the bearing of D from B. Use the unrounded bearing from (b), and give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{15^2+8^2}\)	✓ 1
A1 17 km	✓ 1
Part (b)
M1 \(\tan^{-1}\left(\frac{15}{8}\right)\) or \(90^\circ-\tan^{-1}\left(\frac{8}{15}\right)\)	✓ 1
A1 061.9° (accept 61.9°)	✓ 1
Part (c)
M1 \(61.93^\circ+180^\circ\) (back bearing)	✓ 1
A1 241.9°	✓ 1

Exam tip: A bearing is measured clockwise from north at the point you are travelling from; add or subtract 180° to find a back bearing.

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Calculator

An escalator in a department store rises through a vertical height of 4.5 m. The escalator makes an angle of 30° with the horizontal.

Diagram not to scale

(a)

Calculate the length of the escalator.

(2)

(b)

Calculate the horizontal distance covered by the escalator. Give your answer correct to 3 significant figures.

(2)

(c)

The escalator moves at a constant speed of 0.75 m/s along its slope. Calculate the time taken to travel from the bottom to the top.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\frac{4.5}{\sin30^\circ}\)	✓ 1
A1 9 m	✓ 1
Part (b)
M1 \(9\cos30^\circ\) or \(\frac{4.5}{\tan30^\circ}\)	✓ 1
A1 7.79 m	✓ 1
Part (c)
M1 \(\frac{9}{0.75}\) (ECF from (a))	✓ 1
A1 12 s	✓ 1

Exam tip: Check that your calculator is in degree mode before using sin, cos or tan with an angle in degrees.

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Calculator

The end wall of a carport that carries solar panels is an isosceles triangle ABC with AB = AC = 5 m and BC = 6 m. The point M is the midpoint of BC.

Diagram not to scale

(a)

Show that the height AM of the wall is 4 m.

(3)

(b)

Calculate angle ABC. Give your answer correct to 1 decimal place.

(2)

(c)

Calculate the area of the end wall.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 \(BM=3\) m	✓ 1
M1 \(AM^2=5^2-3^2\)	✓ 1
A1 \(AM=\sqrt{16}=4\) m shown	✓ 1
Part (b)
M1 \(\cos^{-1}\left(\frac{3}{5}\right)\) or \(\tan^{-1}\left(\frac{4}{3}\right)\)	✓ 1
A1 53.1°	✓ 1
Part (c)
M1 \(\frac{1}{2}\times6\times4\)	✓ 1
A1 12 m\(^2\)	✓ 1

Common error: Using the slanted side 5 m as the height of the triangle; the height must be perpendicular to the base.

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Calculator

A beekeeper builds a straight wooden ramp so that bees can be moved on a trolley up to a hive platform. The ramp is 2.4 m long and the top of the ramp is 0.7 m above the ground.

Diagram not to scale

(a)

Calculate the angle the ramp makes with the horizontal ground. Give your answer correct to 1 decimal place.

(2)

(b)

Calculate the horizontal distance from the bottom of the ramp to the point directly below the top of the ramp. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sin θ = 0.7 ÷ 2.4	✓ 1
A1 17.0°	✓ 1
Part (b)
M1 horizontal distance = √(2.4² − 0.7²) (or 2.4 cos θ)	✓ 1
A1 2.30 m	✓ 1

Exam tip: Draw the right-angled triangle and label the hypotenuse first, then choose sin, cos or tan from the sides you know.

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Calculator

A rectangular apiary measures 24 m by 18 m. A beekeeper walks from one corner to the opposite corner.

Diagram not to scale

(a)

Calculate the length of the diagonal of the apiary.

(2)

(b)

The beekeeper first walks along the two sides of the apiary instead. Work out how much further this is than walking along the diagonal.

(2)

(c)

A bee flies along the diagonal at a constant speed of 5 m/s. Calculate the time it takes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 24² + 18² = 900	✓ 1
A1 30 m	✓ 1
Part (b)
M1 24 + 18 = 42 m and 42 − 30	✓ 1
A1 12 m (ECF from (a))	✓ 1
Part (c)
M1 time = 30 ÷ 5 (ECF from (a))	✓ 1
A1 6 s	✓ 1

Common error: adding 24² and 18² and forgetting to take the square root at the end.

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Calculator

A medical clinic C is 12 km from a hospital H on a bearing of 065° from H.

Diagram not to scale

(a)

Calculate how far east of H the clinic is. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate how far north of H the clinic is. Give your answer correct to 3 significant figures.

(2)

(c)

Find the bearing of H from C.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12 sin 65°	✓ 1
A1 10.9 km	✓ 1
Part (b)
M1 12 cos 65°	✓ 1
A1 5.07 km	✓ 1
Part (c)
B1 245°	✓ 1

Exam tip: A bearing is measured clockwise from north and is written with three figures. The bearing back is found by adding or subtracting 180°.

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No calculator

The roof of a beehive has a cross-section in the shape of an isosceles triangle. The base is 48 cm and each sloping side is 30 cm.

Diagram not to scale

(a)

Work out the perpendicular height of the roof.

(2)

(b)

Find the cosine of the angle between a sloping side and the base, as a decimal.

(2)

(c)

Work out the area of the cross-section.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 30² − 24² = 324	✓ 1
A1 18 cm	✓ 1
Part (b)
M1 cos x = 24 ÷ 30	✓ 1
A1 0.8	✓ 1
Part (c)
M1 ½ × 48 × 18 (ECF from (a))	✓ 1
A1 432 cm²	✓ 1

Exam tip: Drawing the height splits an isosceles triangle into two identical right-angled triangles.

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Calculator

A warehouse robot starts at a docking station. It drives 1.2 m due east and then 0.9 m due north, and stops.

Diagram not to scale

(a)

Calculate the straight-line distance from the docking station to the robot's final position.

(2)

(b)

Calculate the angle between the robot's straight-line path back to the station and the eastward direction, to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{1.2^2+0.9^2}\)	✓ 1
A1 1.5 m	✓ 1
Part (b)
M1 \(\tan^{-1}\left(\frac{0.9}{1.2}\right)\)	✓ 1
A1 36.9°	✓ 1

Exam tip: Sketch the right-angled triangle and label the sides opposite, adjacent and hypotenuse before choosing sin, cos or tan.

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Calculator

A fashion designer cuts a right-angled triangular piece of silk. The longest side is 65 cm and one of the shorter sides is 25 cm.

Diagram not to scale

(a)

Calculate the length of the other shorter side.

(2)

(b)

Calculate the area of the piece of silk.

(2)

(c)

Calculate the angle opposite the 25 cm side. Give your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{65^2-25^2}\)	✓ 1
A1 60 cm	✓ 1
Part (b)
M1 \(\frac{1}{2}\times25\times60\)	✓ 1
A1 750 cm²	✓ 1
Part (c)
M1 \(\sin^{-1}\left(\frac{25}{65}\right)\)	✓ 1
A1 22.6°	✓ 1

Common error: Using \(65^2+25^2\) instead of subtracting: the longest side is the hypotenuse, so its square is the sum of the other two squares.

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Calculator

A fishing boat leaves port P and sails 12 km on a bearing of 090°. It then sails 5 km on a bearing of 000° and stops at F.

Diagram not to scale

(a)

Calculate the distance PF.

(2)

(b)

Calculate the bearing of F from P. Give your answer to 1 decimal place.

(2)

(c)

Find the bearing of P from F. Use the unrounded angle from part (b) and give your answer to 1 decimal place.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{12^2+5^2}\)	✓ 1
A1 13 km	✓ 1
Part (b)
M1 \(\tan^{-1}\left(\frac{12}{5}\right)\)	✓ 1
A1 067.4°	✓ 1
Part (c)
A1 247.4° (ECF from (b))	✓ 1

Exam tip: Bearings are measured clockwise from north and written with three figures, so 67.4° is written 067.4°.

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Calculator

A robot stands on level ground 35 m from the foot of a vertical mast. The angle of elevation of the top of the mast from the robot is 28°.

Diagram not to scale

(a)

Calculate the height of the mast. Give your answer to 3 significant figures.

(2)

(b)

A second robot is on the same level ground, 60 m from the foot of the mast. Calculate the angle of elevation of the top of the mast from this robot. Use 18.6 m for the height and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(35\tan28^\circ\)	✓ 1
A1 18.6 m	✓ 1
Part (b)
M1 \(\tan^{-1}\left(\frac{18.6}{60}\right)\)	✓ 1
A1 17.2° (ECF from (a))	✓ 1

Common error: Setting the calculator to radians. Check it shows degrees before using tan, sin or cos on an angle in degrees.

+ Add to test✓ In your test70IGMC3-6-09r23-01Pythagoras' theorem and right-angled trigonometry Paper CEasy4 marks

Calculator

An aircraft is descending in a straight line towards the start of a runway. At point P the aircraft is at an altitude of 2.4 km and its horizontal distance from the start of the runway is 15 km. The runway start is at ground level.

Diagram not to scale

(a)

Calculate the angle of descent, measured below the horizontal. Give your answer correct to 1 decimal place.

(2)

(b)

Calculate the distance the aircraft flies from P to the start of the runway. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of tan with 2.4 and 15, e.g. tan x = 2.4/15	✓ 1
A1 9.1°	✓ 1
Part (b)
M1 Pythagoras, e.g. sqrt(15² + 2.4²) or 2.4/sin(9.09...)	✓ 1
A1 15.2 km	✓ 1

Exam tip: Write the two sides you know next to the angle before choosing sin, cos or tan.

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Calculator

A rectangular field on a farm measures 240 m by 180 m. A farmer builds a straight fence along one diagonal of the field to split it into two triangular paddocks.

Diagram not to scale

(a)

Calculate the length of the fence.

(2)

(b)

A worker walks from one end of the diagonal to the other by going along two sides of the field instead of along the diagonal. Calculate how much further this is than the diagonal.

(2)

(c)

The fence costs €3.25 per metre. Calculate the total cost of the fence.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 240² + 180² = 90 000 or sqrt(240² + 180²)	✓ 1
A1 300 m	✓ 1
Part (b)
B1 total along sides 240 + 180 = 420 m	✓ 1
B1 420 − 300 = 120 m (ECF from (a))	✓ 1
Part (c)
M1 300 × 3.25 (ECF from (a))	✓ 1
A1 €975 (accept 975.00)	✓ 1

Exam tip: Check that the hypotenuse is the longest side before you finish.

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Calculator

A 5.2 m ladder leans against a vertical wall on level ground in an orchard. The foot of the ladder is 2.0 m from the wall.

Diagram not to scale

(a)

Calculate how high up the wall the ladder reaches.

(2)

(b)

Calculate the angle between the ladder and the ground. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5.2² − 2.0² = 23.04	✓ 1
A1 4.8 m	✓ 1
Part (b)
M1 cos x = 2.0/5.2 (or tan x = 4.8/2.0)	✓ 1
A1 67.4°	✓ 1

Common error: When the hypotenuse is unknown add the squares; when a shorter side is unknown subtract.

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Calculator

A wheelchair ramp at a marine research centre rises 0.9 m vertically over a horizontal distance of 10.8 m. The ramp is a straight slope.

Diagram not to scale (vertical scale exaggerated)

(a)

Calculate the length of the ramp. Give your answer to 3 significant figures.

(2)

(b)

Calculate the angle the ramp makes with the horizontal. Give your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{0.9^2 + 10.8^2}\)	✓ 1
A1 10.8 m	✓ 1
Part (b)
M1 \(\tan^{-1}\left(\frac{0.9}{10.8}\right)\) (or equivalent sine or cosine ratio)	✓ 1
A1 4.8°	✓ 1

Common error: Check that your calculator is in degree mode; in radian mode this angle appears as 0.0831.

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Calculator

A research vessel is at point \(A\). A tagged whale is at point \(B\), on a bearing of 065° from \(A\). A second tagged whale is at point \(C\), on a bearing of 155° from \(A\). \(AB = 6\) km and \(AC = 8\) km.

Diagram not to scale

(a)

Find the bearing of \(A\) from \(B\).

(2)

(b)

Show that angle \(BAC = 90°\).

(1)

(c)

Calculate the distance \(BC\).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(65 + 180\) (or co-interior angles used)	✓ 1
A1 245°	✓ 1
Part (b)
B1 \(155 - 65 = 90°\)	✓ 1
Part (c)
M1 \(\sqrt{6^2 + 8^2}\)	✓ 1
A1 10 km	✓ 1

Exam tip: A back bearing differs from the original by 180°: add 180° if the bearing is below 180°, subtract 180° otherwise.

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Calculator

A delivery drone leaves a warehouse roof and climbs in a straight line at 25° to the horizontal. It flies 340 m along this path.

Diagram not to scale

(a)

Calculate the height gained, to the nearest metre.

(2)

(b)

Calculate the horizontal distance covered, to the nearest metre.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(340 \sin 25°\)	✓ 1
A1 144 m	✓ 1
Part (b)
M1 \(340 \cos 25°\)	✓ 1
A1 308 m	✓ 1

Common error: Sine links the angle with the side opposite it: the height is opposite the 25° angle, so use \(\sin\) for the height and \(\cos\) for the horizontal distance.

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Calculator

A camera is lowered from a survey boat on a straight cable that makes an angle of 38° with the horizontal sea surface. The camera is 26 m along the cable from the boat.

Diagram not to scale

(a)

Calculate the depth of the camera below the surface. Give your answer to 3 significant figures.

(2)

(b)

The cable is lengthened, keeping the same angle, so that the camera is 25 m below the surface. Calculate the length of cable that is now out. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(26 \sin 38°\)	✓ 1
A1 16.0 m	✓ 1
Part (b)
M1 \(\frac{25}{\sin 38°}\)	✓ 1
A1 40.6 m	✓ 1

Exam tip: When the unknown is the hypotenuse, the sine ratio appears in the denominator: length \(= \frac{\text{opposite}}{\sin\theta}\).

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Calculator

A bridge has a vertical tower that is 40 m high. A straight cable runs from the top of the tower to an anchor point on level ground, 55 m from the base of the tower.

Diagram not to scale

(a)

Calculate the length of the cable. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the angle that the cable makes with the ground. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of Pythagoras, 40² + 55² = 4625	✓ 1
A1 68.0 m	✓ 1
Part (b)
M1 tan x = 40 ÷ 55	✓ 1
A1 36.0°	✓ 1

Exam tip: Check that the angle you find is sensible: a cable that is shorter vertically than horizontally should be below 45° to the ground.

+ Add to test✓ In your test78IGMC3-6-08r26-02Pythagoras' theorem and right-angled trigonometry Paper CEasy5 marks

No calculator

A rectangular solar panel on a space probe is 1.2 m long and 0.5 m wide. The diagonal of the panel makes an angle x with the longer side.

Diagram not to scale

(a)

Show that the length of the diagonal is 1.3 m.

(2)

(b)

Write down sin x as a fraction in its simplest form.

(2)

(c)

Write down cos x as a fraction in its simplest form.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 1.2² + 0.5² = 1.44 + 0.25 = 1.69	✓ 1
A1 √1.69 = 1.3 m, shown	✓ 1
Part (b)
M1 sin x = 0.5 ÷ 1.3	✓ 1
A1 5/13	✓ 1
Part (c)
B1 12/13 (= 0.923 to 3 significant figures)	✓ 1

Common error: Sine uses the side opposite the angle, and cosine the side next to it. The longer side is next to angle x.

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Calculator

A tracking camera is on level ground, 4.5 km horizontally from the launch pad of a rocket. The rocket rises vertically from the pad.

Diagram not to scale

(a)

When the rocket is 2.8 km above the ground, calculate the angle of elevation of the rocket from the camera. Give your answer correct to 1 decimal place.

(2)

(b)

Calculate the distance from the camera to the rocket at this time.

(2)

(c)

Later, the angle of elevation of the rocket from the camera is 60°. Calculate the height of the rocket above the ground at this time. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan x = 2.8 ÷ 4.5	✓ 1
A1 31.9°	✓ 1
Part (b)
M1 √(4.5² + 2.8²) = √28.09	✓ 1
A1 5.30 km (accept 5.3)	✓ 1
Part (c)
M1 height = 4.5 × tan 60°	✓ 1
A1 7.79 km	✓ 1

Common error: Do not use the distance from the camera to the rocket as the adjacent side: the adjacent side is the horizontal distance on the ground.

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Calculator

The frame of a small coffee roaster is a triangle with sides of length 0.8 m, 1.5 m and 1.7 m.

Diagram not to scale

(a)

Show that the triangle is right-angled.

(2)

(b)

Calculate the smallest angle of the triangle. Give your answer correct to 1 decimal place.

(2)

(c)

The frame is made from steel tube that costs €6.40 per metre. Calculate the cost of the tube for one frame.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.8² + 1.5² = 0.64 + 2.25 = 2.89	✓ 1
A1 1.7² = 2.89, so Pythagoras holds and the triangle is right-angled	✓ 1
Part (b)
M1 sin x = 0.8 ÷ 1.7 (accept cos or tan with correct sides)	✓ 1
A1 28.1°	✓ 1
Part (c)
M1 perimeter 0.8 + 1.5 + 1.7 = 4.0 m	✓ 1
A1 €25.60	✓ 1

Common error: The smallest angle is opposite the shortest side, so use 0.8 m as the opposite side.

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Calculator

An aircraft takes off and climbs in a straight line at a constant angle of 12° to the horizontal. It flies a distance of 5.4 km along this straight path.

Diagram not to scale

(a)

Calculate the horizontal distance the aircraft has travelled.

(2)

(b)

Calculate the height the aircraft has gained.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 horizontal distance = 5.4 × cos 12°	✓ 1
A1 5.28 km	✓ 1
Part (b)
M1 height = 5.4 × sin 12°	✓ 1
A1 1.12 km	✓ 1

Common error: Check that your calculator is in degree mode, and match each side to the angle: the side opposite the angle uses sine, the side next to it uses cosine.

+ Add to test✓ In your test82IGMC3-6-11r27-02Sine rule, cosine rule, trigonometric graphs and 3D problems Paper CEasy4 marks

No calculator

A farmer has a rectangular field that measures 40 m by 30 m. A fence is to be built round the edge of the field and also along one diagonal of the field. Fencing costs €12 per metre.

Diagram not to scale

(a)

Work out the length of the diagonal of the field.

(2)

(b)

Work out the total cost of the fencing.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 40² + 30² (= 2500)	✓ 1
A1 50 m	✓ 1
Part (b)
M1 total length = 2(40 + 30) + 50 (= 190 m)	✓ 1
A1 €2280	✓ 1

Exam tip: Read what is being fenced: the question asks for the edge and the diagonal, so add the perimeter and the diagonal before multiplying by the cost per metre.

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Calculator

An aircraft leaves airport A and flies 120 km on a bearing of 070° to airport B. It then flies 90 km on a bearing of 160° to airport C.

Diagram not to scale

(a)

Show that angle ABC = 90°.

(2)

(b)

Calculate the distance AC.

(2)

(c)

Calculate the bearing of C from A. Give your answer to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 bearing of A from B is 250° (or 70° + 180°)	✓ 1
A1 250° − 160° = 90°	✓ 1
Part (b)
M1 AC² = 120² + 90² (= 22 500)	✓ 1
A1 150 km	✓ 1
Part (c)
M1 angle BAC = tan⁻¹(90 ÷ 120)	✓ 1
A1 36.9°	✓ 1
A1 bearing = 70° + 36.87…° = 106.9°	✓ 1

Common error: A bearing is measured clockwise from north, so add the angle BAC to the bearing of B from A rather than subtracting it; sketching the north lines first shows which is correct.

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Calculator

A stage engineer fixes a cable from the top of a vertical pole to a point on level ground. The pole is 6.5 m tall and the point on the ground is 8.0 m from the foot of the pole. The cable is straight.

Diagram not to scale

(a)

Calculate the length of the cable.

(2)

(b)

Calculate the angle the cable makes with the ground.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sqrt(6.5² + 8.0²)	✓ 1
A1 10.3 m	✓ 1
Part (b)
M1 tan⁻¹(6.5 ÷ 8.0)	✓ 1
A1 39.1°	✓ 1

Exam tip: Write down which two sides you know (here opposite and adjacent) before choosing tan, sin or cos.

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Calculator

A vertical radio mast on a bridge is 45 m tall. A straight steel cable runs from the top of the mast to a point on the level deck that is 28 m from the foot of the mast.

(a)

Calculate the length of the cable.

(2)

(b)

Calculate the angle between the cable and the deck. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{45^2+28^2}\)	✓ 1
A1 53.0 m	✓ 1
Part (b)
M1 \(\tan^{-1}\left(\frac{45}{28}\right)\)	✓ 1
A1 58.1°	✓ 1

Exam tip: Label the opposite, adjacent and hypotenuse sides relative to the angle before choosing sin, cos or tan.

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No calculator

A rectangular lid on a coffee roasting drum measures 24 cm by 7 cm. The diagonal of the lid makes an angle \(x\) with the longer side.

(a)

Work out the length of the diagonal.

(2)

(b)

Write down the exact value of \(\sin x\) and of \(\cos x\) as fractions.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{24^2+7^2}=\sqrt{625}\)	✓ 1
A1 25 cm	✓ 1
Part (b)
B1 \(\sin x=\frac{7}{25}\)	✓ 1
B1 \(\cos x=\frac{24}{25}\)	✓ 1

Common error: Sine is opposite over hypotenuse; the side opposite angle x is the shorter side here.

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Calculator

A conveyor belt that carries green coffee beans into a roaster is 3.5 m long and rises at an angle of 18° to the horizontal.

(a)

Calculate the height gained by the beans along the belt. Give your answer correct to 3 significant figures.

(2)

(b)

A new belt must lift the beans through a height of 1.5 m at the same angle of 18°. Calculate the length of the new belt. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(3.5\sin 18°\)	✓ 1
A1 1.08 m	✓ 1
Part (b)
M1 \(\frac{1.5}{\sin 18°}\)	✓ 1
A1 4.85 m	✓ 1

Exam tip: Make sure the calculator is in degree mode before using sin, cos or tan.

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Calculator

A boom on a space probe is 2.6 m long. It rests with its upper end against a vertical wall at a height of 2.0 m above level ground, and its lower end on the ground.

(a)

Calculate the horizontal distance from the foot of the wall to the lower end of the boom. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the angle between the boom and the ground. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{2.6^2-2.0^2}\)	✓ 1
A1 1.66 m	✓ 1
Part (b)
M1 \(\sin^{-1}\left(\frac{2.0}{2.6}\right)\)	✓ 1
A1 50.3°	✓ 1

Exam tip: When two sides are known, choose the trigonometric ratio that uses only given sides, not a rounded answer.

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Calculator

A wheelchair ramp at the entrance of a pharmacy is a straight slope 3.6 m long. Its top is 0.45 m above the horizontal pavement.

(a)

Calculate the angle between the ramp and the horizontal. Give your answer correct to 1 decimal place.

(2)

(b)

Calculate the horizontal length of the ramp. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sin x = 0.45/3.6 (or equivalent)	✓ 1
A1 7.2° (7.18...)	✓ 1
Part (b)
M1 √(3.6² − 0.45²) or 3.6 cos(7.18...°)	✓ 1
A1 3.57 m (accept 3.57 from using 7.2°)	✓ 1

Exam tip: Label which side is opposite, adjacent and hypotenuse before choosing sin, cos or tan.

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Calculator

On a flat square-grid map in a video game, a character starts at a base B, walks 90 m due north and then 120 m due east to reach a point C.

(a)

Calculate the straight-line distance BC.

(2)

(b)

Calculate the bearing of C from B. Give your answer correct to 1 decimal place.

(2)

(c)

Write down the bearing of B from C, correct to 1 decimal place. Use your answer to (b).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(90² + 120²) or 90² + 120² = 22500	✓ 1
A1 150 m	✓ 1
Part (b)
M1 tan θ = 120/90 (angle from north)	✓ 1
A1 053.1°	✓ 1
Part (c)
A1 233.1° (ECF: 180 + answer to (b))	✓ 1

Common error: Bearings are measured clockwise from north and written with three figures, so 53.1° should be written as 053.1°.

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Calculator

A triangular warning label on a medicine pack has sides of length 7.5 cm, 10 cm and 12.5 cm.

(a)

Show that the label is a right-angled triangle.

(2)

(b)

Calculate the smallest angle of the label. Give your answer correct to 1 decimal place.

(2)

(c)

Calculate the area of the label.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 7.5² + 10² = 156.25	✓ 1
A1 12.5² = 156.25, so Pythagoras' theorem holds and the triangle is right-angled	✓ 1
Part (b)
M1 sin x = 7.5/12.5 or tan x = 7.5/10	✓ 1
A1 36.9°	✓ 1
Part (c)
A1 ½ × 7.5 × 10 = 37.5 cm²	✓ 1

Exam tip: To show a triangle is right-angled, test whether the squares of the two shorter sides add to the square of the longest side.

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Calculator

A rectangular gaming monitor has a screen of width 60 cm and height 34 cm.

(a)

Calculate the length of the diagonal of the screen. Give your answer correct to 3 significant figures. Keep the unrounded value for part (b).

(2)

(b)

Screens are advertised by their diagonal length in inches. Given that 1 inch = 2.54 cm, calculate the diagonal in inches, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(60² + 34²)	✓ 1
A1 69.0 cm	✓ 1
Part (b)
M1 68.96... ÷ 2.54	✓ 1
A1 27.2 inches	✓ 1

Exam tip: Convert units as the last step, using the unrounded value, to avoid carrying a rounding error into the answer.

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Calculator

A delivery van leaves a depot D and drives 18 km in a straight line on a bearing of 070° to a warehouse W. Give distances to 3 significant figures.

Diagram not to scale

(a)

Write down the bearing of D from W.

(1)

(b)

Calculate how far north of D the warehouse W is.

(2)

(c)

Calculate how far east of D the warehouse W is.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 250°	✓ 1
Part (b)
M1 18 × cos 70°	✓ 1
A1 6.16 km	✓ 1
Part (c)
M1 18 × sin 70°	✓ 1
A1 16.9 km	✓ 1

Common error: Bearings are measured clockwise from north, so the north component uses cosine and the east component uses sine of the bearing.

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No calculator

The gable end of a building is an isosceles triangle. The two equal sloping sides are each 13 m long and the horizontal base is 24 m long.

Diagram not to scale

(a)

Work out the perpendicular height of the gable end.

(3)

(b)

Work out the area of the gable end.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 half of base = 12 m	✓ 1
M1 13² − 12² (= 169 − 144)	✓ 1
A1 5 m	✓ 1
Part (b)
M1 ½ × 24 × 5 (ECF from (a))	✓ 1
A1 60 m²	✓ 1

Exam tip: Splitting an isosceles triangle down the middle gives two congruent right-angled triangles.

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Calculator

A fixed rope on a mountain face runs in a straight line from a base camp on level ground to a ledge. The ledge is 45 m vertically above the ground and 60 m horizontally from the camp.

Diagram not to scale

(a)

Calculate the length of the rope.

(2)

(b)

A second straight rope of length 85 m reaches a ledge that is 36 m above the ground. Calculate the horizontal distance from the camp to this ledge.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 attempt to use Pythagoras, \(\sqrt{45^2+60^2}\)	✓ 1
A1 75 m	✓ 1
Part (b)
M1 \(\sqrt{85^2-36^2}\) or \(85^2-36^2=5929\)	✓ 1
A1 77 m	✓ 1

Common error: When finding a shorter side, subtract the squares; adding them gives the hypotenuse.

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Calculator

A telecoms mast stands on level ground and is 28 m tall. A surveyor stands on the ground 35 m from the base of the mast.

Diagram not to scale

(a)

Calculate the angle of elevation of the top of the mast from the surveyor. Give your answer to 1 decimal place.

(2)

(b)

A straight support cable runs from the top of the mast to the ground and makes an angle of 62° with the ground. Calculate the length of the cable. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\tan\theta=\dfrac{28}{35}\)	✓ 1
A1 38.7°	✓ 1
Part (b)
M1 \(\sin 62°=\dfrac{28}{L}\) or \(L=\dfrac{28}{\sin 62°}\)	✓ 1
A1 31.7 m	✓ 1

Exam tip: Label the opposite, adjacent and hypotenuse relative to the angle before choosing sin, cos or tan.

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Calculator

A food technologist uses a cone-shaped mould to make chocolate. The mould has a circular top of radius 5 cm and a vertical height of 12 cm.

Cone with its circular top uppermost. Diagram not to scale

(a)

Calculate the slant height of the cone.

(2)

(b)

Calculate the angle between the slant edge and the vertical axis of the cone. Give your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{5^2+12^2}\)	✓ 1
A1 13 cm	✓ 1
Part (b)
M1 \(\tan x=\dfrac{5}{12}\) (accept \(\sin x=\dfrac{5}{13}\))	✓ 1
A1 22.6°	✓ 1

Common error: The angle with the vertical axis is opposite the radius, not the height, so use radius over height for tangent.

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Calculator

A fibre-optic cable is laid in a straight line up a hillside. The cable is 120 m long and makes an angle of 18° with the horizontal.

Diagram not to scale

(a)

Calculate the vertical height gained by the cable. Give your answer to 3 significant figures.

(2)

(b)

Calculate the horizontal distance covered by the cable. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(120\sin 18°\)	✓ 1
A1 37.1 m	✓ 1
Part (b)
M1 \(120\cos 18°\)	✓ 1
A1 114 m	✓ 1

Exam tip: Check that your calculator is in degree mode before using sin, cos or tan with angles in degrees.

+ Add to test✓ In your test99IGMC3-6-11r40-01Pythagoras' theorem and right-angled trigonometry Paper CEasy4 marks

Calculator

A planetary lander has a vertical antenna mast of height 24 cm standing on level ground. A support wire runs from the top of the mast to a ground anchor that is 18 cm from the foot of the mast.

Diagram not to scale

(a)

Calculate the length of the support wire.

(2)

(b)

Calculate the angle between the wire and the ground. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{24^2+18^2}\) or \(\sqrt{900}\)	✓ 1
A1 30 cm	✓ 1
Part (b)
M1 \(\tan\theta=\frac{24}{18}\) or \(\sin\theta=\frac{24}{30}\) or \(\cos\theta=\frac{18}{30}\)	✓ 1
A1 53.1°	✓ 1

Common error: Writing 24 + 18 = 42 as the wire length, or using 24² − 18². The wire is the hypotenuse, so the squares are added.

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Calculator

A research boat leaves harbour H and sails 15 km on a bearing of 040° to a buoy B.

Diagram not to scale

(a)

Calculate how far east of H the buoy is. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate how far north of H the buoy is. Give your answer correct to 3 significant figures.

(2)

(c)

Find the bearing of H from B.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 15 sin 40°	✓ 1
A1 9.64 km	✓ 1
Part (b)
M1 15 cos 40°	✓ 1
A1 11.5 km	✓ 1
Part (c)
B1 220°	✓ 1

Exam tip: A bearing is measured clockwise from north, so the north component uses cosine of the bearing and the east component uses sine, for angles up to 90°.

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Calculator

A rover leaves a lander along a straight ramp. The ramp is 4.5 m long and is inclined at 15° to the horizontal.

Diagram not to scale

(a)

Calculate the vertical height of the top of the ramp above the ground. Give your answer correct to 3 significant figures.

(2)

(b)

A second ramp is 3.6 m long and its top is 0.9 m above the ground. Calculate the angle this ramp makes with the horizontal. Give your answer correct to 1 decimal place.

(2)

(c)

State, with a reason, which ramp is steeper.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 4.5 sin 15°	✓ 1
A1 1.16 m	✓ 1
Part (b)
M1 \(\sin\theta=\frac{0.9}{3.6}\)	✓ 1
A1 14.5°	✓ 1
Part (c)
B1 the first ramp, because 15° is greater than 14.5°	✓ 1

Exam tip: Check that your calculator is in degree mode. A common slip is to get 0.0 or a tiny value when it is in radian mode.

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Calculator

Archaeologists mark out a rectangular trench 7.5 m long and 4.0 m wide. A measuring tape is stretched from one corner of the trench to the opposite corner.

Diagram not to scale

(a)

Calculate the length of the tape.

(2)

(b)

Calculate the angle between the tape and a long side of the trench. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 7.5² + 4.0² (= 72.25)	✓ 1
A1 8.5 m	✓ 1
Part (b)
M1 tan⁻¹(4.0 ÷ 7.5) (or sin⁻¹(4.0 ÷ 8.5), cos⁻¹(7.5 ÷ 8.5))	✓ 1
A1 28.1°	✓ 1

Exam tip: In a right-angled triangle, choose sin, cos or tan by naming the sides first: opposite, adjacent and hypotenuse.

+ Add to test✓ In your test103IGMC3-6-10r45-02Pythagoras' theorem and right-angled trigonometry Paper CEasy5 marks

No calculator

A cyclist rides up a straight ramp of length 13 m. The ramp rises 5 m vertically. The ramp makes an angle x with the horizontal.

Diagram not to scale

(a)

Calculate the horizontal length of the ramp.

(2)

(b)

Write down the value of tan x as a fraction.

(1)

(c)

A second ramp makes the same angle x with the horizontal and rises 3.0 m. Calculate its horizontal length.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 13² − 5² (= 169 − 25 = 144)	✓ 1
A1 12 m	✓ 1
Part (b)
B1 5/12	✓ 1
Part (c)
M1 3.0 ÷ (5/12) (or 3.0 × 12/5, or similar triangles with scale factor 0.6)	✓ 1
A1 7.2 m	✓ 1

Exam tip: Pythagorean triples such as 5, 12, 13 and 3, 4, 5 save time in no-calculator questions.

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Calculator

An archaeologist leans a ladder of length 6.5 m against the vertical wall of a trench. The foot of the ladder is on level ground 2.5 m from the wall.

Diagram not to scale

(a)

Calculate how high up the wall the ladder reaches.

(2)

(b)

The top of the ladder slips down the wall by 0.8 m. The ladder is still 6.5 m long and its foot is still on the ground. Calculate how far the foot of the ladder has moved away from the wall.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6.5² − 2.5² (= 36)	✓ 1
A1 6.0 m	✓ 1
Part (b)
B1 new height = 6.0 − 0.8 = 5.2 m (ECF from (a))	✓ 1
M1 √(6.5² − 5.2²) (= 3.9)	✓ 1
A1 3.9 − 2.5 = 1.4 m	✓ 1

Common error: When a side other than the hypotenuse is needed, subtract the squares; adding them gives a length longer than the hypotenuse.

+ Add to test✓ In your test105IGMC3-6-10r45-06Bearings Paper CEasy5 marks

Calculator

A cyclist rides 18 km due north and then 24 km due east.

Diagram not to scale

(a)

Calculate the distance of the cyclist from the starting point.

(2)

(b)

Calculate the bearing of the finishing point from the starting point. Give your answer correct to 1 decimal place.

(2)

(c)

Find the bearing of the starting point from the finishing point. Give your answer correct to 1 decimal place.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 18² + 24² (= 900)	✓ 1
A1 30 km	✓ 1
Part (b)
M1 tan⁻¹(24 ÷ 18)	✓ 1
A1 053.1°	✓ 1
Part (c)
B1 053.1° + 180° = 233.1° (ECF from (b))	✓ 1

Exam tip: Bearings are measured clockwise from north and written with three figures, for example 053.1°.

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Calculator

A beekeeper leans a ladder of length 4.5 m against a vertical wall to reach a platform. The foot of the ladder is on level ground 1.2 m from the wall.

Diagram not to scale

(a)

Calculate how high up the wall the ladder reaches. Give your answer to 3 significant figures.

(3)

(b)

Calculate the angle the ladder makes with the ground. Give your answer to 1 decimal place. Use the lengths given in the question.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 height² = 4.5² − 1.2²	✓ 1
M1 = 18.81	✓ 1
A1 4.34 m	✓ 1
Part (b)
M1 cos θ = 1.2 ÷ 4.5	✓ 1
A1 74.5°	✓ 1

Common error: Use Pythagoras with a subtraction when the unknown side is a shorter side: height² = hypotenuse² − base².

+ Add to test✓ In your test107IGMC3-6-12r45-02Pythagoras' theorem and right-angled trigonometry Paper CEasy4 marks

Calculator

A wheelchair ramp at a hospital entrance rises 0.85 m over a horizontal distance of 6.8 m. The ramp surface is a straight slope.

Diagram not to scale (vertical scale exaggerated)

(a)

Calculate the angle the ramp makes with the horizontal. Give your answer to 1 decimal place.

(2)

(b)

Calculate the length of the sloping surface of the ramp. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan θ = 0.85 ÷ 6.8	✓ 1
A1 7.1°	✓ 1
Part (b)
M1 length² = 0.85² + 6.8²	✓ 1
A1 6.85 m	✓ 1

Exam tip: Choose the trigonometric ratio from the sides you have: here you are given opposite and adjacent, so use tangent.

+ Add to test✓ In your test108IGMC3-6-12r45-11Pythagoras' theorem and right-angled trigonometry Paper CEasy5 marks

No calculator

A rectangular pit garage floor has length 24 m and a diagonal of length 25 m.

Diagram not to scale

(a)

Calculate the width of the floor.

(2)

(b)

Calculate the perimeter of the floor.

(2)

(c)

Write down the value of the sine of the angle between the diagonal and the length of 24 m, as a fraction.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 width² = 25² − 24² = 625 − 576 = 49	✓ 1
A1 7 m	✓ 1
Part (b)
M1 2 × (24 + 7) (ECF from (a))	✓ 1
A1 62 m	✓ 1
Part (c)
B1 7/25 (ECF from (a))	✓ 1

Exam tip: Without a calculator, look for whole-number Pythagorean triples such as 7, 24, 25.

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Calculator

A medical helicopter hovers 85 m directly above a point on level ground. The angle of depression of a landing marker from the helicopter is 24°.

Diagram not to scale

(a)

Calculate the horizontal distance from the point to the marker. Give your answer to 3 significant figures.

(3)

(b)

Calculate the direct distance from the helicopter to the marker. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 the angle at the marker in the right-angled triangle is 24° (alternate angles)	✓ 1
M1 85 ÷ tan 24°	✓ 1
A1 191 m	✓ 1
Part (b)
M1 85 ÷ sin 24°	✓ 1
A1 209 m	✓ 1

Common error: The angle of depression is measured down from the horizontal at the helicopter; it equals the angle of elevation from the marker.

+ Add to test✓ In your test110IGMC3-6-11r46-01Bearings Paper CEasy6 marks

Calculator

A marine biologist sails a research boat 8.4 km due north from harbour H and then 6.3 km due east to reach a survey point S.

Diagram not to scale

(a)

Calculate the distance HS.

(2)

(b)

Calculate the bearing of S from H, giving your answer to 1 decimal place.

(2)

(c)

Find the bearing of H from S, giving your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of Pythagoras, \(\sqrt{8.4^2+6.3^2}\)	✓ 1
A1 10.5 km	✓ 1
Part (b)
M1 \(\tan\theta=\frac{6.3}{8.4}\) with the angle measured from north	✓ 1
A1 036.9°	✓ 1
Part (c)
M1 adds 180° to the bearing in (b) (ECF from (b))	✓ 1
A1 216.9°	✓ 1

Common error: A bearing is measured clockwise from north and is written with three figures, so 36.9° must be written as 036.9°.

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Calculator

A museum entrance has a straight access ramp. The ramp rises 0.75 m over a horizontal distance of 9.0 m.

Diagram not to scale (vertical scale exaggerated)

(a)

Calculate the angle the ramp makes with the horizontal. Give your answer to 1 decimal place.

(2)

(b)

Calculate the length of the sloping surface of the ramp. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\tan\theta=\frac{0.75}{9.0}\)	✓ 1
A1 4.8°	✓ 1
Part (b)
M1 \(\sqrt{9.0^2+0.75^2}\) (or \(0.75\div\sin\) of the angle from the unrounded angle)	✓ 1
A1 9.03 m	✓ 1

Exam tip: Draw a quick right-angled triangle and label the sides opposite, adjacent and hypotenuse before choosing sin, cos or tan.

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Calculator

A logistics company has two depots. Depot B is 50 km from depot A on a bearing of 072°.

Diagram not to scale

(a)

Write down the bearing of A from B.

(1)

(b)

Calculate how far north of A the depot B is. Give your answer to 3 significant figures.

(2)

(c)

Calculate how far east of A the depot B is. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 252°	✓ 1
Part (b)
M1 \(50\cos72^\circ\)	✓ 1
A1 15.5 km	✓ 1
Part (c)
M1 \(50\sin72^\circ\)	✓ 1
A1 47.6 km	✓ 1

Common error: The northward distance uses cosine of the bearing and the eastward distance uses sine, because the bearing is measured from north.

+ Add to test✓ In your test113IGMC3-6-11r46-07Pythagoras' theorem and right-angled trigonometry Paper CEasy5 marks

Calculator

A ladder of length 5.2 m rests against a vertical wall. The foot of the ladder is on level ground 2.0 m from the base of the wall.

Diagram not to scale

(a)

Calculate the height on the wall that the top of the ladder reaches.

(2)

(b)

The safety guidance says the angle between the ladder and the ground should be at least 70°. Calculate this angle for the ladder, to 1 decimal place, and state whether the guidance is met.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{5.2^2-2.0^2}\)	✓ 1
A1 4.8 m	✓ 1
Part (b)
M1 \(\cos\theta=\frac{2.0}{5.2}\)	✓ 1
A1 67.4°	✓ 1
B1 not met because \(67.4<70\)	✓ 1

Common error: When finding a side by Pythagoras, subtract the squares if you are finding a shorter side; adding them gives the hypotenuse only.

+ Add to test✓ In your test114IGMC-046-01-05Bearings Paper CMedium6 marks

No calculator

The bearing of Y from X is 112°. The bearing of Z from Y is 230°.

Diagram not to scale

(a)

Find the bearing of X from Y.

(1)

(b)

Find the size of angle XYZ.

(2)

(c)

XY = YZ. Find the size of angle YXZ.

(1)

(d)

Find the bearing of Z from X.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 292°	✓ 1
Part (b)
M1: 292 − 230	✓ 1
A1: 62°	✓ 1
Part (c)
B1: (180 − 62) ÷ 2 = 59°	✓ 1
Part (d)
M1: 112 + their angle YXZ	✓ 1
A1: 171°	✓ 1

Exam tip: Draw a north line at every point where a bearing is measured. Angles between two bearings taken at the same point can then be found by subtraction.

+ Add to test✓ In your test115IGMC-046-01-08Bearings Paper CMedium6 marks

Calculator

Two coastguard stations, A and B, are on a straight east–west coastline. B is 15 km due east of A. A boat C is out at sea, north of the coastline. The bearing of C from A is 040° and the bearing of C from B is 320°.

Diagram not to scale

(a)

Find angle CAB and angle CBA.

(2)

(b)

Calculate how far north of the coastline the boat is.

(2)

(c)

Calculate the distance AC.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: angle CAB = 90 − 40 = 50°	✓ 1
B1: angle CBA = 320 − 270 = 50°	✓ 1
Part (b)
M1: using the isosceles triangle, so the foot of the perpendicular from C is the midpoint of AB: height = 7.5 × tan 50°, or any equivalent complete method, e.g. AC = 15 sin 50° ÷ sin 80° then height = AC × sin 50°	✓ 1
A1: 8.94 km (accept 8.938…)	✓ 1
Part (c)
M1: AC = 7.5 ÷ cos 50°, or any equivalent complete method, e.g. AC = 15 sin 50° ÷ sin 80°, or AC = height ÷ sin 50°, or Pythagoras with 7.5 and the height	✓ 1
A1: 11.7 km (accept 11.66 to 11.67)	✓ 1

Exam tip: When the two base angles are equal, the triangle is isosceles. The perpendicular from C then meets AB at its midpoint, which gives two right-angled triangles, each with a base of 7.5 km.

+ Add to test✓ In your test116IGMC-046-01-10Bearings Paper CMedium5 marks

No calculator

A drone flies 6 km from P on a bearing of 075° to Q. It then flies 8 km from Q on a bearing of 165° to R.

Diagram not to scale

(a)

Find the bearing of P from Q.

(1)

(b)

Show that angle PQR is a right angle.

(2)

(c)

Find the distance PR.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 255°	✓ 1
Part (b)
M1: 255 − 165	✓ 1
A1: = 90°	✓ 1
Part (c)
M1: 6² + 8² = 100	✓ 1
A1: 10 km	✓ 1

Exam tip: In a 'show that' question, every step must be written down. Give the back bearing 255° explicitly before subtracting.

+ Add to test✓ In your test117IGMC-046-00-04Bearings Paper CMedium6 marks

Calculator

A ship leaves harbour H and sails 35 km due north. It then sails 18 km due east to a point S.
Give distances correct to 3 significant figures and bearings correct to 1 decimal place.

Diagram not to scale

(a)

Calculate the distance HS.

(3)

(b)

Calculate the bearing of S from H.

(2)

(c)

Find the bearing of H from S.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 35² + 18² (= 1549)	✓ 1
M1: √1549	✓ 1
A1: 39.4 km (39.357…)	✓ 1
Part (b)
M1: tan θ = 18 ÷ 35, or sin θ = 18 ÷ 39.4, or cos θ = 35 ÷ 39.4, where θ is the angle at H measured from north	✓ 1
A1: 027.2° (accept 27.2°)	✓ 1
Part (c)
B1: 207.2° (follow through: their bearing + 180°)	✓ 1

Common error: Using tan θ = 35 ÷ 18 gives the angle measured from east, not from north. Bearings are always measured clockwise from north.

+ Add to test✓ In your test118IGMC-046-00-06Pythagoras' theorem and right-angled trigonometry Paper CMedium7 marks

Calculator

An isosceles triangle has two equal sides of length 11 cm and a base of length 8 cm.
Give your answers correct to 3 significant figures, or to 1 decimal place for angles.

Diagram not to scale

(a)

Calculate the perpendicular height of the triangle.

(3)

(b)

Calculate the size of one of the base angles.

(2)

(c)

Calculate the area of the triangle.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: half the base is 4 cm	✓ 1
M1: √(11² − 4²) = √105	✓ 1
A1: 10.2 cm (10.246…)	✓ 1
Part (b)
M1: cos θ = 4 ÷ 11, or an equivalent such as sin θ = √105 ÷ 11 or tan θ = √105 ÷ 4	✓ 1
A1: 68.7° (68.676…)	✓ 1
Part (c)
M1: ½ × 8 × their height, or ½ × 11 × 11 × sin(180° − 2 × their angle)	✓ 1
A1: 41.0 cm² (40.987…), using the unrounded height. Using 10.2 gives 40.8, which is not accurate enough for this mark	✓ 1

Common error: The height splits the base in half, so use 4 cm, not 8 cm, in the right-angled triangle. Keep the full calculator value of the height when you find the area. Rounding to 10.2 first gives 40.8 cm², which is not accurate to 3 significant figures.

+ Add to test✓ In your test119IGMC-046-00-07Pythagoras' theorem and right-angled trigonometry Paper CMedium6 marks

Calculator

Ravi stands on level ground, 45 m from the base of a vertical tower. His eyes are 1.6 m above the ground. The angle of elevation from his eyes to the top of the tower is 32°.

Diagram not to scale

(a)

Calculate the height of the tower.

(3)

(b)

A car is parked on the ground, 70 m from the base of the tower. Calculate the angle of depression of the car from the top of the tower.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: tan 32° = x ÷ 45, where x is the height of the top of the tower above Ravi's eye level	✓ 1
M1: 45 tan 32° + 1.6	✓ 1
A1: 29.7 m (29.71…)	✓ 1
Part (b)
M1: right-angled triangle with opposite = their height of tower and adjacent = 70, e.g. tan θ = 29.7… ÷ 70	✓ 1
M1: θ = tan⁻¹(29.7… ÷ 70)	✓ 1
A1: 23.0°	✓ 1

Common error: Trigonometry gives only the height above Ravi's eye level. Remember to add the 1.6 m to find the height of the tower.

+ Add to test✓ In your test120IGMC-046-00-08Pythagoras' theorem and right-angled trigonometry Paper CMedium6 marks

Calculator

A rectangular garden is 15.4 m long and 9.2 m wide. Two straight paths run along the diagonals of the garden.

Diagram not to scale

(a)

Calculate the length of one diagonal.

(2)

(b)

Calculate the angle between a diagonal and a longer side of the garden.

(2)

(c)

Calculate the acute angle between the two diagonal paths.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: √(15.4² + 9.2²) or √321.8	✓ 1
A1: 17.9 m (17.938…)	✓ 1
Part (b)
M1: tan θ = 9.2 ÷ 15.4, or an equivalent correct sin or cos ratio using the diagonal	✓ 1
A1: 30.9° (30.85…)	✓ 1
Part (c)
M1: 2 × their angle between a diagonal and a longer side, or 180° − 2 × (90° − their angle), or another complete correct method such as the cosine rule in a triangle formed by the diagonals	✓ 1
A1: 61.7° (accept 61.8° from using a rounded angle of 30.9°)	✓ 1

Exam tip: The diagonals of a rectangle bisect each other, so they form isosceles triangles. Use the unrounded angle from your calculator when doubling it, to avoid rounding errors in the final answer.

+ Add to test✓ In your test121IGMC-046-00-09Pythagoras' theorem and right-angled trigonometry Paper CMedium6 marks

Calculator

Triangle ABC has a right angle at B and AB = 8.5 cm.
D is a point on BC, between B and C.
Angle ADB = 55° and angle ACB = 28°.

Diagram not to scale

(a)

Calculate BD.

(2)

(b)

Calculate BC.

(2)

(c)

Hence find DC.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 8.5 ÷ tan 55° or 8.5 × tan 35°	✓ 1
A1: 5.95 cm (5.951…)	✓ 1
Part (b)
M1: 8.5 ÷ tan 28° or 8.5 × tan 62°	✓ 1
A1: 16.0 cm (15.986…)	✓ 1
Part (c)
M1: their BC − their BD	✓ 1
A1: 10.0 cm (10.03…); accept 10.1 from correctly rounded earlier answers (16.0 − 5.95)	✓ 1

Exam tip: Both triangles share the side AB, so AB is the opposite side in each. Use tan in each triangle and keep full calculator values until the end to avoid rounding errors in the final subtraction.

+ Add to test✓ In your test122IGMC3-6-03d-03Bearings Paper CMedium5 marks

Calculator

A ferry leaves a port and travels 36 km on a bearing of 040° in a straight line to an island. Use unrounded values in your working.

Diagram not to scale

(a)

Calculate how far north of the port the island is. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate how far east of the port the island is. Give your answer correct to 3 significant figures.

(2)

(c)

Find the bearing of the port from the island.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 northward distance = 36 cos 40°	✓ 1
A1 27.6 km	✓ 1
Part (b)
M1 eastward distance = 36 sin 40°	✓ 1
A1 23.1 km	✓ 1
Part (c)
B1 220°	✓ 1

Common error: Bearings are measured clockwise from north, so the angle in a right-angled triangle is measured from the north line, not from the east line.

+ Add to test✓ In your test123IGMC3-6-03d-06Sine rule, cosine rule, trigonometric graphs and 3D problems Paper CMedium5 marks

Calculator

The lamp of a lighthouse is 48 m above sea level. Two boats, A and B, are in line with the lighthouse and on the same side of it. From the lamp, the angle of depression of A is 15° and the angle of depression of B is 32°. Use unrounded values in your working.

Diagram not to scale

(a)

Calculate the horizontal distance from the foot of the lighthouse to A.

(2)

(b)

Calculate the horizontal distance from the foot of the lighthouse to B.

(2)

(c)

Calculate the distance AB. Give your answer correct to the nearest metre.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan 15° = 48 ÷ d_A (angle of depression equals the angle of elevation from the boat)	✓ 1
A1 179.1 m	✓ 1
Part (b)
M1 tan 32° = 48 ÷ d_B	✓ 1
A1 76.8 m	✓ 1
Part (c)
B1 102 m (ECF from (a) and (b): d_A − d_B)	✓ 1

Exam tip: The angle of depression from the top is equal to the angle of elevation from the bottom, because they are alternate angles.

+ Add to test✓ In your test124IGMC3-6-03d-07Bearings Paper CMedium7 marks

Calculator

A hiker walks 6.0 km due east and then 3.5 km due south, where she stops. Use unrounded values in your working.

Diagram not to scale

(a)

Calculate the straight-line distance from the start to the stopping point. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the bearing of the stopping point from the start. Give your answer correct to 1 decimal place.

(4)

(c)

Find the bearing of the start from the stopping point. Give your answer correct to 1 decimal place.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(6.0² + 3.5²)	✓ 1
A1 6.95 km	✓ 1
Part (b)
M1 tan θ = 3.5 ÷ 6.0, where θ is the angle below the east direction	✓ 1
A1 θ = 30.3° (30.3°)	✓ 1
M1 bearing = 90° + θ	✓ 1
A1 120.3°	✓ 1
Part (c)
B1 300.3° (ECF from (b): add 180°)	✓ 1

Common error: A bearing is always measured clockwise from north; do not stop at the angle inside the triangle.

+ Add to test✓ In your test125IGMC3-6-03d-10Pythagoras' theorem and right-angled trigonometry Paper CMedium7 marks

Calculator

The end wall of a roof truss is an isosceles triangle. The base is 8.4 m long and the two equal sloping sides are each 5.0 m long.

Diagram not to scale

(a)

Calculate the vertical height of the triangle. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate the angle each sloping side makes with the base. Give your answer correct to 1 decimal place.

(2)

(c)

Calculate the area of the triangle. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 half of base = 4.2 m, forming a right-angled triangle with hypotenuse 5.0 m	✓ 1
M1 h = √(5.0² − 4.2²) = √7.36	✓ 1
A1 2.71 m	✓ 1
Part (b)
M1 cos θ = 4.2 ÷ 5.0 (or tan θ = 2.71 ÷ 4.2)	✓ 1
A1 32.9°	✓ 1
Part (c)
M1 area = ½ × 8.4 × 2.71 (ECF from (a))	✓ 1
A1 11.4 m²	✓ 1

Exam tip: In an isosceles triangle, split it down the middle to make two right-angled triangles. The perpendicular height then bisects the base.

+ Add to test✓ In your test126IGMC3-6-10r5-05Bearings Paper CMedium7 marks

Calculator

In a sailing game, a ship leaves port A and sails 12 km on a bearing of 040° to B. It then sails 9 km on a bearing of 130° to C.

Diagram not to scale

(a)

Show that angle ABC is 90°.

(2)

(b)

Calculate the distance AC.

(2)

(c)

Calculate the bearing of C from A, to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 the bearing of A from B is 040° + 180° = 220°	✓ 1
A1 220° − 130° = 90°	✓ 1
Part (b)
M1 √(12² + 9²)	✓ 1
A1 15 km	✓ 1
Part (c)
M1 tan(BAC) = 9/12	✓ 1
A1 angle BAC = 36.9°	✓ 1
A1 bearing 76.9°	✓ 1

Exam tip: Mark north lines at each point on a sketch, then use parallel lines to find the angles between legs.

+ Add to test✓ In your test127IGMC3-6-10r5-06Pythagoras' theorem and right-angled trigonometry Paper CMedium5 marks

Calculator

A wheelchair ramp at a pharmacy rises 0.45 m over a horizontal distance of 5.4 m.

Diagram not to scale

(a)

Calculate the angle the ramp makes with the horizontal, to 1 decimal place.

(2)

(b)

Calculate the length of the sloping surface of the ramp, to 3 significant figures.

(2)

(c)

A rule says the ramp angle must not exceed 5°. State whether this ramp meets the rule, with a reason.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan θ = 0.45/5.4	✓ 1
A1 4.8°	✓ 1
Part (b)
M1 √(5.4² + 0.45²)	✓ 1
A1 5.42 m	✓ 1
Part (c)
B1 yes, because 4.8° is less than 5° (ECF from (a))	✓ 1

Common error: Use the inverse function tan⁻¹ to find an angle; applying tan to a ratio gives a number, not an angle.

+ Add to test✓ In your test128IGMC3-6-10r5-07Pythagoras' theorem and right-angled trigonometry Paper CMedium6 marks

Calculator

In a video game, a player stands 26 m from the base of a vertical tower, on level ground, and sees the top at an angle of elevation of 38°. Use unrounded values in your working and give answers to 3 significant figures.

Diagram not to scale

(a)

Calculate the height of the tower.

(2)

(b)

The player walks directly towards the tower until the angle of elevation is 55°. Calculate the new distance from the base of the tower.

(2)

(c)

Calculate the distance the player has walked.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 26 × tan 38°	✓ 1
A1 20.3 m	✓ 1
Part (b)
M1 20.3 ÷ tan 55° (unrounded height accepted)	✓ 1
A1 14.2 m	✓ 1
Part (c)
M1 26 − 14.2 (use unrounded value)	✓ 1
A1 11.8 m	✓ 1

Common error: Rounding the tower height early can shift the final answer; carry the full calculator value forward.

+ Add to test✓ In your test129IGMC3-6-10r5-08Bearings Paper CMedium5 marks

Calculator

A drone leaves a hospital pharmacy H and flies 5 km due north to X, then 12 km due east to Y.

Diagram not to scale

(a)

Calculate the distance HY.

(2)

(b)

Calculate the bearing of Y from H, to 1 decimal place.

(2)

(c)

Find the bearing of H from Y, to 1 decimal place.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(5² + 12²)	✓ 1
A1 13 km	✓ 1
Part (b)
M1 tan⁻¹(12/5)	✓ 1
A1 067.4°	✓ 1
Part (c)
B1 247.4° (ECF from (b))	✓ 1

Exam tip: Write bearings with three figures; 67.4° is written 067.4°.

+ Add to test✓ In your test130IGMC3-6-09r6-04Pythagoras' theorem and right-angled trigonometry Paper CMedium4 marks

Calculator

A touring band uses a straight ramp to roll a harp case onto a stage. The ramp is 3.2 m long and the stage edge is 0.85 m above the ground.

Diagram not to scale

(a)

Calculate the angle the ramp makes with the ground. Give your answer to 3 significant figures.

(2)

(b)

At another venue the stage edge is 1.1 m above the ground. The band wants the ramp at the same angle as before. Calculate the length of ramp needed. Use the unrounded angle and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses sin θ = 0.85/3.2	✓ 1
θ = 15.4°	✓ 1
Part (b)
Uses length = 1.1/sin θ (or ratio 1.1/0.85 × 3.2)	✓ 1
Length = 4.14 m	✓ 1

Exam tip: If the angle is unchanged, the ratio height : length is unchanged, so a scale factor gives a quick check.

+ Add to test✓ In your test131IGMC3-6-09r6-05Bearings Paper CMedium5 marks

Calculator

A rectangular wheat field measures 150 m by 80 m. A farmer walks from one corner straight across the field to the opposite corner.

Diagram not to scale

(a)

Calculate the length of the diagonal path.

(2)

(b)

The farmer could instead walk along two sides of the field. Calculate how much shorter the diagonal path is.

(1)

(c)

Calculate the angle between the diagonal and a 150 m side. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses Pythagoras: 150² + 80² = 28 900	✓ 1
Diagonal = 170 m	✓ 1
Part (b)
150 + 80 = 230 m along the sides, so the saving is 230 − 170 = 60 m	✓ 1
Part (c)
Uses tan θ = 80/150	✓ 1
θ = 28.1°	✓ 1

Common error: Do not forget to take the square root after adding the squares; 28 900 is not a length.

+ Add to test✓ In your test132IGMC3-6-09r6-11Bearings Paper CMedium5 marks

Calculator

A passenger aircraft flies 150 km on a bearing of 040° and then 150 km on a bearing of 130°. Treat the ground as flat.

Diagram not to scale

(a)

Show that the angle between the two legs of the journey is 90°.

(1)

(b)

Calculate the distance of the aircraft from its starting point. Give your answer to 3 significant figures.

(2)

(c)

Find the bearing of the starting point from the aircraft's final position.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
130° − 40° = 90° between the directions of travel, so the legs meet at a right angle	✓ 1
Part (b)
Pythagoras: 150² + 150² = 45 000	✓ 1
Distance = 212 km	✓ 1
Part (c)
The triangle is right-angled and isosceles so the base angles are 45°	✓ 1
Reverse bearing of the second leg is 310°, so the bearing of the start is 310° − 45° = 265°	✓ 1

Exam tip: Sketch a north line at each turning point; angles between a bearing and a north line are alternate or co-interior with those at the previous point.

+ Add to test✓ In your test133IGMC3-6-09r6-14Bearings Paper CMedium6 marks

Calculator

An airliner is 1.8 km vertically above a runway when the pilot sees the runway threshold at an angle of depression of 7°. Assume level ground.

Diagram not to scale

(a)

Calculate the horizontal distance from the airliner to the threshold. Give your answer to 3 significant figures.

(2)

(b)

Calculate the straight-line distance from the airliner to the threshold. Give your answer to 3 significant figures.

(2)

(c)

The airliner flies along this line at a constant 240 km/h. Calculate the time taken to reach the threshold, in minutes. Use the unrounded distance and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Angle of elevation from threshold = 7° (alternate angles)	✓ 1
tan 7° = 1.8/distance, so distance = 14.7 km	✓ 1
Part (b)
Uses sin 7° = 1.8/distance (or Pythagoras with their answer to (a))	✓ 1
Distance = 14.8 km	✓ 1
Part (c)
Time = distance/speed = 14.770/240 hours, ECF from (b)	✓ 1
Time = 3.69 minutes	✓ 1

Common error: The angle of depression is measured from the horizontal at the airliner, not from the vertical.

+ Add to test✓ In your test134IGMC3-6-10r6-04Bearings Paper CMedium7 marks

Calculator

A harbour H is 14 km from a lighthouse L. The bearing of L from H is 072°. An island I is 10.5 km from L on a bearing of 162° from L.

Diagram not to scale

(a)

Write down the bearing of H from L.

(1)

(b)

Show that angle HLI = 90°.

(1)

(c)

Calculate the distance HI.

(2)

(d)

Calculate the bearing of I from H. Give your answer to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
72° + 180° = 252°	✓ 1
Part (b)
252° − 162° = 90°	✓ 1
Part (c)
Pythagoras: 14² + 10.5² = 306.25	✓ 1
HI = 17.5 km	✓ 1
Part (d)
Angle LHI: tan LHI = 10.5/14	✓ 1
Angle LHI = 36.9°	✓ 1
I lies clockwise of L as seen from H, so bearing = 72° + 36.9° = 108.9°	✓ 1

Common error: Add the angle to the bearing of L only if I lies clockwise (to the right) of L as seen from H; sketch it to check the direction.

+ Add to test✓ In your test135IGMC3-6-10r6-05Pythagoras' theorem and right-angled trigonometry Paper CMedium5 marks

Calculator

A spotlight is fixed to a bar 6.5 m above the stage floor. An actor's head is 1.7 m above the floor and 3.6 m horizontally from the spotlight.

Diagram not to scale

(a)

Calculate the vertical distance between the spotlight and the actor's head.

(1)

(b)

Calculate the distance from the spotlight to the actor's head.

(2)

(c)

Calculate the angle of depression of the actor's head from the spotlight. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
6.5 − 1.7 = 4.8 m	✓ 1
Part (b)
Pythagoras: 4.8² + 3.6² = 36	✓ 1
Distance = 6.0 m	✓ 1
Part (c)
tan θ = 4.8/3.6 (angle below the horizontal)	✓ 1
θ = 53.1°	✓ 1

Exam tip: An angle of depression is measured downwards from the horizontal through the observer.

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A volcano has its vent at V. Seismic station A is 7.2 km due east of V and station B is 9.6 km due north of V.

Diagram not to scale

(a)

Calculate the distance AB.

(2)

(b)

Calculate the bearing of B from A. Give your answer to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Pythagoras: 7.2² + 9.6² = 144	✓ 1
AB = 12.0 km	✓ 1
Part (b)
B is north-west of A; acute angle with north: tan θ = 7.2/9.6 so θ = 36.9°	✓ 1
Bearing = 360° − 36.9°	✓ 1
Bearing = 323.1°	✓ 1

Common error: A bearing in the north-west quadrant lies between 270° and 360°; do not leave the answer as the acute angle.

+ Add to test✓ In your test137IGMC3-6-10r6-08Bearings Paper CMedium5 marks

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A fishing boat leaves a harbour and sails 35 km on a bearing of 200°.

Diagram not to scale

(a)

Write down the bearing of the harbour from the boat.

(1)

(b)

Calculate how far south of the harbour the boat is. Give your answer to 3 significant figures.

(2)

(c)

Calculate how far west of the harbour the boat is. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
200° − 180° = 020°	✓ 1
Part (b)
Angle between the course and south is 20°; south distance = 35 cos 20°	✓ 1
South distance = 32.9 km	✓ 1
Part (c)
West distance = 35 sin 20°	✓ 1
West distance = 12.0 km	✓ 1

Common error: Bearing 200° is 20° past south, not 20° past north; draw the north line and measure clockwise.

+ Add to test✓ In your test138IGMC3-6-09r7-03Bearings Paper CMedium7 marks

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A cyclist leaves a campsite and rides 18 km on a bearing of 070° to a village. She then rides 12 km on a bearing of 160° to a lake.

Diagram not to scale

(a)

Show that the angle between the two stages of the ride, at the village, is 90°.

(2)

(b)

Calculate the direct distance from the campsite to the lake. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the bearing of the lake from the campsite. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of the campsite from the village is 070° + 180° = 250°	✓ 1
A1 250° − 160° = 90°	✓ 1
Part (b)
M1 use of Pythagoras, 18² + 12²	✓ 1
A1 21.6 km	✓ 1
Part (c)
M1 tan (angle at campsite) = 12 ÷ 18	✓ 1
A1 angle at campsite = 33.7°	✓ 1
A1 bearing = 070° + 33.7° = 103.7°	✓ 1

Common error: A bearing is measured clockwise from north. The angle found inside the triangle must be added to 070°, not given as the bearing itself.

+ Add to test✓ In your test139IGMC3-6-09r7-09Bearings Paper CMedium5 marks

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A rectangular park measures 0.9 km by 1.2 km. One cyclist rides along two adjacent sides of the park. Another cyclist rides along the straight path across the park from one corner to the opposite corner. Both cyclists travel at a constant speed of 15 km/h.

Diagram not to scale

(a)

Calculate the length of the straight path across the park.

(2)

(b)

Calculate how many minutes less the second cyclist takes.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 0.9² + 1.2² = 2.25	✓ 1
A1 1.5 km	✓ 1
Part (b)
M1 difference in distance = (0.9 + 1.2) − 1.5 = 0.6 km	✓ 1
M1 0.6 ÷ 15 × 60	✓ 1
A1 2.4 minutes	✓ 1

Common error: Convert hours to minutes at the end: 0.04 hours is 2.4 minutes, not 0.04 minutes or 4 minutes.

+ Add to test✓ In your test140IGMC3-6-09r7-10Sine rule, cosine rule, trigonometric graphs and 3D problems Paper CMedium5 marks

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A road up a hill has a constant gradient of 8%. This means the road rises 8 m vertically for every 100 m travelled horizontally.

Diagram not to scale

(a)

Calculate the angle the road makes with the horizontal. Give your answer correct to 1 decimal place.

(2)

(b)

A cyclist climbs a total height of 150 m. Calculate the distance she rides along the road. Use the unrounded angle from (a) and give your answer correct to the nearest metre.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan x = 8 ÷ 100	✓ 1
A1 4.6°	✓ 1
Part (b)
M1 distance = 150 ÷ sin x	✓ 1
M1 substitution of the unrounded angle (tan⁻¹ 0.08)	✓ 1
A1 1881 m	✓ 1

Exam tip: A percentage gradient is a tangent ratio written as a percentage. Convert it to tan x = 0.08 before using the inverse function.

+ Add to test✓ In your test141IGMC3-6-09r7-11Bearings Paper CMedium6 marks

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The finishing line of a charity cycle race is 14.0 km due east and 9.5 km due south of the starting line.

Diagram not to scale

(a)

Calculate the straight-line distance from the start to the finish. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the bearing of the finish from the start. Give your answer correct to 1 decimal place.

(3)

(c)

Write down the bearing of the start from the finish, correct to 1 decimal place.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of Pythagoras, 14.0² + 9.5²	✓ 1
A1 16.9 km	✓ 1
Part (b)
M1 tan (angle south of east) = 9.5 ÷ 14.0	✓ 1
A1 angle = 34.2°	✓ 1
A1 bearing = 090° + 34.2° = 124.2°	✓ 1
Part (c)
B1 124.2° + 180° = 304.2°, ECF from (b)	✓ 1

Common error: Starting from east, the angle measured towards south must be added to 090°, because bearings are measured clockwise from north.

+ Add to test✓ In your test142IGMC3-6-10r7-04Bearings Paper CMedium5 marks

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A sea swimmer leaves a beach and swims 400 m in a straight line on a bearing of 235°.

Diagram not to scale

(a)

Calculate how far south of the starting point the swimmer is. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate how far west of the starting point the swimmer is. Give your answer correct to 3 significant figures.

(2)

(c)

Write down the bearing of the starting point from the swimmer.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 angle from south is 235° − 180° = 55°, so distance south = 400 × cos 55°	✓ 1
A1 229 m	✓ 1
Part (b)
M1 distance west = 400 × sin 55°	✓ 1
A1 328 m	✓ 1
Part (c)
B1 235° − 180° = 055°	✓ 1

Common error: The angle in the triangle is measured from the south line, 235° − 180° = 55°, not from north. Using 235° directly in sin or cos gives the wrong lengths.

+ Add to test✓ In your test143IGMC3-6-10r7-07Pythagoras' theorem and right-angled trigonometry Paper CMedium6 marks

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A support cable is attached to a vertical mast 42 m above level ground. The other end of the cable is fixed to the ground 18 m from the foot of the mast.

Diagram not to scale

(a)

Calculate the length of the cable. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the angle between the cable and the ground. Give your answer correct to 1 decimal place.

(2)

(c)

A second cable, 54 m long, is fixed to the same point on the mast and to the ground. Calculate its distance from the foot of the mast. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 42² + 18² = 2088	✓ 1
A1 45.7 m	✓ 1
Part (b)
M1 tan x = 42 ÷ 18	✓ 1
A1 66.8°	✓ 1
Part (c)
M1 54² − 42² = 1152	✓ 1
A1 33.9 m	✓ 1

Common error: When the hypotenuse is known and a shorter side is needed, subtract the squares. Adding 54² and 42² is a frequent slip.

+ Add to test✓ In your test144IGMC3-6-10r7-08Sine rule, cosine rule, trigonometric graphs and 3D problems Paper CMedium6 marks

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Three turbines A, B and C are on level ground. The bearing of B from A is 064°. The bearing of C from B is 150°. AB = BC.

Diagram not to scale

(a)

Show that angle ABC = 94°.

(2)

(b)

Find angle BAC.

(2)

(c)

Find the bearing of C from A.

(1)

(d)

Find the bearing of A from C.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of A from B is 064° + 180° = 244°	✓ 1
A1 244° − 150° = 94°	✓ 1
Part (b)
M1 triangle is isosceles, so angle BAC = (180° − 94°) ÷ 2	✓ 1
A1 43°	✓ 1
Part (c)
B1 064° + 43° = 107°, ECF from (b)	✓ 1
Part (d)
B1 107° + 180° = 287°, ECF from (c)	✓ 1

Exam tip: Mark the north line at each point on your sketch. Parallel north lines let you find angles inside the triangle using co-interior or alternate angles.

+ Add to test✓ In your test145IGMC3-6-10r7-14Pythagoras' theorem and right-angled trigonometry Paper CMedium6 marks

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A library ladder 4.5 m long rests against a vertical bookshelf on level ground. At first the foot of the ladder is 1.5 m from the shelf. The foot is then pulled 0.9 m further away from the shelf, so that it is 2.4 m from the shelf.

Diagram not to scale

(a)

Calculate the height reached by the top of the ladder at first. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the height reached by the top of the ladder after the foot is pulled out. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate how far the top of the ladder has moved down the shelf. Use the unrounded heights and give your answer correct to 2 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 4.5² − 1.5² = 18	✓ 1
A1 4.24 m	✓ 1
Part (b)
M1 4.5² − 2.4² = 14.49	✓ 1
A1 3.81 m	✓ 1
Part (c)
M1 difference of the two heights, √18 − √14.49	✓ 1
A1 0.44 m	✓ 1

Common error: Do not assume the top of the ladder moves down by the same 0.9 m that the foot moves out. The two movements are linked by Pythagoras, not equal.

+ Add to test✓ In your test146IGMC3-6-10r8-05Bearings Paper CMedium7 marks

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A hiker walks from P to Q, 8 km on a bearing of 040°. She then walks 6 km from Q to R on a bearing of 130°.

Diagram not to scale

(a)

Show that angle PQR = 90°.

(2)

(b)

Calculate the distance PR.

(2)

(c)

Calculate the bearing of R from P. Give your answer to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of P from Q = 040° + 180° = 220°	✓ 1
A1 220° − 130° = 90°	✓ 1
Part (b)
M1 √(8² + 6²)	✓ 1
A1 10 km	✓ 1
Part (c)
M1 angle QPR = tan⁻¹(6/8) = 36.87°	✓ 1
M1 40° + 36.87°	✓ 1
A1 76.9°	✓ 1

Exam tip: Draw a north line at each point; angles on a straight north-south line and alternate angles then give the bearings.

+ Add to test✓ In your test147IGMC3-6-10r8-06Bearings Paper CMedium6 marks

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A telecoms drone leaves base B and flies 5 km due east to X. It then flies 3 km due north to Y. Give bearings to 1 decimal place.

Diagram not to scale

(a)

Calculate the distance BY. Give your answer to 3 significant figures.

(2)

(b)

Calculate the bearing of Y from B.

(3)

(c)

Write down the bearing of B from Y.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(5² + 3²)	✓ 1
A1 5.83 km	✓ 1
Part (b)
M1 angle XBY = tan⁻¹(3/5)	✓ 1
M1 90° − angle XBY	✓ 1
A1 59.0°	✓ 1
Part (c)
B1 239.0°	✓ 1

Common error: Giving angle XBY as the bearing; a bearing is measured clockwise from north.

+ Add to test✓ In your test148IGMC3-6-10r8-07Pythagoras' theorem and right-angled trigonometry Paper CMedium4 marks

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A mountaineer stands at the top of a vertical cliff 85 m high. A boat on the sea below is seen at an angle of depression of 22°.

Diagram not to scale

(a)

Calculate the horizontal distance from the foot of the cliff to the boat. Give your answer to 3 significant figures.

(2)

(b)

The boat later moves so that it is 150 m from the foot of the cliff. Calculate the new angle of depression. Give your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 85 ÷ tan 22°	✓ 1
A1 210 m	✓ 1
Part (b)
M1 tan⁻¹(85/150)	✓ 1
A1 29.5°	✓ 1

Common error: Measuring the angle of depression from the vertical; it is measured from the horizontal, and equals the angle of elevation from the boat.

+ Add to test✓ In your test149IGMC3-6-10r8-08Pythagoras' theorem and right-angled trigonometry Paper CMedium6 marks

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A vertical telecoms mast is 24 m tall. Wire 1 runs from the top of the mast to a point on the ground 10 m from its base. Wire 2 runs from a point 15 m up the mast to a point on the ground 8 m from its base, on the opposite side.

Diagram not to scale

(a)

Calculate the length of wire 1.

(2)

(b)

Calculate the length of wire 2.

(2)

(c)

The wire costs €3.20 per metre. Calculate the total cost of both wires.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(24² + 10²)	✓ 1
A1 26 m	✓ 1
Part (b)
M1 √(15² + 8²)	✓ 1
A1 17 m	✓ 1
Part (c)
M1 3.20 × (26 + 17)	✓ 1
A1 €137.60	✓ 1

Exam tip: Recognise Pythagorean triples such as 5, 12, 13 and 8, 15, 17 to save time.

+ Add to test✓ In your test150IGMC3-6-09r9-05Pythagoras' theorem and right-angled trigonometry Paper CMedium5 marks

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A bridge truss is shaped like an isosceles triangle. Its horizontal base is 14 m long and each of the two sloping steel members is 12.5 m long.

Diagram not to scale

(a)

Calculate the height of the truss. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate the angle between a sloping member and the base. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 half the base = 7 m, right-angled triangle	✓ 1
M1 h² = 12.5² − 7² (= 107.25)	✓ 1
A1 10.4 m	✓ 1
Part (b)
M1 cos x = 7 ÷ 12.5 (ECF from (a) if tan or sin used with 10.4)	✓ 1
A1 55.9°	✓ 1

Common error: Using the full base of 14 m in the right-angled triangle. Split the isosceles triangle in half first.

+ Add to test✓ In your test151IGMC3-6-09r9-06Pythagoras' theorem and right-angled trigonometry Paper CMedium6 marks

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An observer on flat ground watches a probe hovering 90 m directly above a landing marker. The angle of elevation of the probe from the observer is 32°.

Diagram not to scale

(a)

Calculate the horizontal distance from the observer to the marker. Give your answer correct to 3 significant figures.

(3)

(b)

The observer walks 60 m directly towards the marker. Use 144 m as the original distance. Calculate the new angle of elevation of the probe. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan 32° = 90 ÷ d	✓ 1
M1 d = 90 ÷ tan 32°	✓ 1
A1 144 m	✓ 1
Part (b)
M1 new distance 144 − 60 = 84 m	✓ 1
M1 tan x = 90 ÷ 84	✓ 1
A1 47.0°	✓ 1

Exam tip: Draw a sketch with the height as the vertical side before choosing sin, cos or tan.

+ Add to test✓ In your test152IGMC3-6-09r9-07Bearings Paper CMedium7 marks

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A delivery drone flies from a depot D on a bearing of 042° for 15 km to a town A. It then flies on a bearing of 132° for 20 km to a farm B.

Diagram not to scale

(a)

Show that angle DAB = 90°.

(2)

(b)

Calculate the distance DB.

(2)

(c)

Find the bearing of B from D. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 bearing of D from A is 222°	✓ 1
B1 222° − 132° = 90°	✓ 1
Part (b)
M1 DB² = 15² + 20²	✓ 1
A1 25 km	✓ 1
Part (c)
M1 tan(angle ADB) = 20 ÷ 15	✓ 1
A1 53.1°	✓ 1
A1 bearing 095.1° (42 + 53.1)	✓ 1

Common error: Using 132° directly as the angle at A. Work out the angle inside the triangle using the back bearing.

+ Add to test✓ In your test153IGMC3-6-09r9-11Sine rule, cosine rule, trigonometric graphs and 3D problems Paper CMedium5 marks

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A coffee roaster's hopper is a hollow cone with its point at the bottom. The circular rim has radius 0.48 m and the vertical height of the cone is 0.90 m.

Cross-section, not to scale

(a)

Calculate the slant height of the cone. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate the angle between the slant side and the vertical axis of the cone. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 l² = 0.48² + 0.90²	✓ 1
M1 l² = 1.0404	✓ 1
A1 1.02 m	✓ 1
Part (b)
M1 tan x = 0.48 ÷ 0.90	✓ 1
A1 28.1°	✓ 1

Exam tip: Check which angle is asked for. The angle with the vertical axis uses the radius as the opposite side.

+ Add to test✓ In your test154IGMC3-6-09r10-06Pythagoras' theorem and right-angled trigonometry Paper CMedium5 marks

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A surveyor stands at point P on level ground, 15 m from the base of a vertical grain silo. The angle of elevation of the top of the silo from P is 54°. Ignore the height of the surveyor.

Diagram not to scale

(a)

Calculate the height of the silo. Give your answer to 3 significant figures.

(2)

(b)

The surveyor walks directly away from the silo to a point Q where the angle of elevation of the top is 35°. Calculate the distance from Q to the base of the silo. Use your unrounded height from (a) and give your answer to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan 54° = h/15	✓ 1
A1 h = 20.6 m	✓ 1
Part (b)
M1 tan 35° = h/d	✓ 1
M1 d = 20.645… ÷ tan 35°	✓ 1
A1 29.5 m	✓ 1

Exam tip: Keep unrounded values in your calculator between parts and round only the final answer.

+ Add to test✓ In your test155IGMC3-6-09r10-07Bearings Paper CMedium5 marks

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A crop-spraying drone takes off from a point O, flies 24 km due south, and then 7 km due west to point S.

Diagram not to scale

(a)

Calculate the straight-line distance OS.

(2)

(b)

Find the bearing of S from O. Give your answer to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 OS² = 24² + 7²	✓ 1
A1 25 km	✓ 1
Part (b)
M1 tan x = 7/24, where x is the angle at O between south and OS	✓ 1
A1 x = 16.3°	✓ 1
A1 bearing = 180° + 16.3° = 196.3°	✓ 1

Common error: A bearing is measured clockwise from north, so an angle measured from the south line must be added to 180°.

+ Add to test✓ In your test156IGMC3-6-09r10-10Pythagoras' theorem and right-angled trigonometry Paper CMedium7 marks

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The frame of a small harp is a right-angled triangle. The vertical pillar is 1.05 m long and the horizontal soundboard is 0.68 m long. The third side is the neck.

Diagram not to scale

(a)

Calculate the length of the neck. Give your answer to 3 significant figures.

(2)

(b)

Calculate the angle between the neck and the soundboard. Give your answer to 1 decimal place.

(2)

(c)

Timber for the frame costs €14.50 per metre. Calculate the cost of timber for all three sides, using the unrounded neck length. Give your answer in euros to 2 decimal places.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 neck² = 1.05² + 0.68²	✓ 1
A1 1.25 m	✓ 1
Part (b)
M1 tan x = 1.05/0.68	✓ 1
A1 57.1°	✓ 1
Part (c)
M1 perimeter = 1.05 + 0.68 + 1.2509…	✓ 1
M1 cost = 2.980… × 14.50	✓ 1
A1 €43.22	✓ 1

Exam tip: For a right-angled triangle, label the sides opposite, adjacent and hypotenuse before choosing sin, cos or tan.

+ Add to test✓ In your test157IGMC3-6-09r11-02Pythagoras' theorem and right-angled trigonometry Paper CMedium5 marks

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A volcanologist stands on level ground 1.2 km (1200 m) horizontally from the point directly below the rim of a crater. The angle of elevation of the rim from her eye is 28°. Ignore the height of her eye above the ground.

Diagram not to scale

(a)

Calculate the height of the rim above the ground. Give your answer to the nearest metre.

(2)

(b)

She walks 300 m directly away from the volcano. Use 638 m for the height. Calculate the new angle of elevation of the rim. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 height = 1200 × tan 28°	✓ 1
A1 638 m	✓ 1
Part (b)
B1 new horizontal distance = 1500 m	✓ 1
M1 tan⁻¹(638 ÷ 1500)	✓ 1
A1 23.0°	✓ 1

Common error: After moving, the horizontal distance becomes 1200 + 300, not 300.

+ Add to test✓ In your test158IGMC3-6-09r11-03Bearings Paper CMedium7 marks

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A ship leaves port P and sails 40 km on a bearing of 060° to a buoy A. It then sails 30 km on a bearing of 150° to a lighthouse B.

Diagram not to scale

(a)

Show that angle PAB = 90°.

(2)

(b)

Calculate the distance PB.

(2)

(c)

Calculate the bearing of P from B. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of P from A = 060° + 180° = 240°	✓ 1
A1 240° − 150° = 90°	✓ 1
Part (b)
M1 PB² = 40² + 30²	✓ 1
A1 50 km	✓ 1
Part (c)
M1 angle APB = tan⁻¹(30 ÷ 40) = 36.9°	✓ 1
M1 bearing of B from P = 060° + 36.9° = 096.9°	✓ 1
A1 bearing of P from B = 276.9° (accept 277°)	✓ 1

Common error: To find a bearing of P from B, add or subtract 180° from the bearing of B from P.

+ Add to test✓ In your test159IGMC3-6-09r11-06Bearings Paper CMedium6 marks

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A ferry leaves harbour X and sails 24 km on a bearing of 072° to harbour Y.

Diagram not to scale

(a)

Calculate how far east of X harbour Y is. Give your answer correct to 1 decimal place.

(2)

(b)

Calculate how far north of X harbour Y is. Give your answer correct to 2 decimal places.

(2)

(c)

Write down the bearing of X from Y.

(1)

(d)

A lifeboat sails due south from Y to a point Z that is due east of X. Write down the distance YZ, correct to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 24 × sin 72°	✓ 1
A1 22.8 km	✓ 1
Part (b)
M1 24 × cos 72°	✓ 1
A1 7.42 km	✓ 1
Part (c)
B1 252°	✓ 1
Part (d)
B1 7.42 km	✓ 1

Exam tip: Bearings are measured clockwise from north, so the east component uses sine and the north component uses cosine.

+ Add to test✓ In your test160IGMC3-6-09r11-13Pythagoras' theorem and right-angled trigonometry Paper CMedium7 marks

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A technician leans a ladder of length 5.2 m against a vertical wall at the side of a stage. The foot of the ladder is on horizontal ground, 2.0 m from the wall.

Diagram not to scale

(a)

Calculate the height on the wall that the top of the ladder reaches.

(2)

(b)

Calculate the angle the ladder makes with the ground. Give your answer correct to 1 decimal place.

(2)

(c)

The foot of the ladder is pulled outwards so that the top slides down the wall by 0.4 m. Calculate the new distance of the foot from the wall. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sqrt(5.2² − 2.0²)	✓ 1
A1 4.8 m	✓ 1
Part (b)
M1 cos⁻¹(2.0 ÷ 5.2)	✓ 1
A1 67.4°	✓ 1
Part (c)
B1 new height = 4.4 m	✓ 1
M1 sqrt(5.2² − 4.4²)	✓ 1
A1 2.77 m	✓ 1

Common error: When you rearrange Pythagoras' theorem for a shorter side, subtract the squares; do not add them.

+ Add to test✓ In your test161IGMC3-6-08r14-03Pythagoras' theorem and right-angled trigonometry Paper CMedium4 marks

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A pharmacy technician folds a circular filter paper into a cone. The cone has slant height 9.0 cm and its circular rim has diameter 8.0 cm.

Cross-section, not to scale

(a)

Calculate the vertical height of the cone, to 3 significant figures.

(2)

(b)

Calculate the angle between the slant height and the vertical axis of the cone, to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 h² = 9.0² − 4.0²	✓ 1
A1 8.06 cm	✓ 1
Part (b)
M1 sin x = 4.0/9.0 (or cos x = 8.06/9.0, or tan x = 4.0/8.06)	✓ 1
A1 26.4°	✓ 1

Exam tip: The radius is half the diameter: halve the 8.0 cm before using it in the right-angled triangle.

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Water pumping stations A, B and C are on level ground. B is 8.0 km from A on a bearing of 040°. C is 6.0 km from B on a bearing of 130°. Give lengths to 3 significant figures and angles to 1 decimal place.

Diagram not to scale

(a)

Show that angle ABC = 90°.

(2)

(b)

Calculate AC.

(2)

(c)

Calculate the bearing of C from A.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 the bearing of A from B is 040° + 180° = 220°	✓ 1
B1 angle ABC = 220° − 130° = 90°	✓ 1
Part (b)
M1 AC² = 8.0² + 6.0²	✓ 1
A1 10.0 km	✓ 1
Part (c)
M1 tan(angle BAC) = 6.0/8.0	✓ 1
A1 angle BAC = 36.9°	✓ 1
A1 bearing 076.9°	✓ 1

Exam tip: Always draw a north line at each point and mark the angles measured clockwise from it.

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On a game map, towers B and C are each 250 m from the base A. B is on a bearing of 070° from A and C is on a bearing of 150° from A. Give lengths to 3 significant figures and angles to 1 decimal place.

Diagram not to scale

(a)

Write down the size of angle BAC.

(1)

(b)

Calculate the distance BC.

(3)

(c)

Calculate the bearing of C from B.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 80°	✓ 1
Part (b)
B1 triangle ABC is isosceles; half of angle BAC = 40°	✓ 1
M1 BC/2 = 250 sin 40° (or equivalent)	✓ 1
A1 321 m	✓ 1
Part (c)
B1 bearing of A from B = 250°	✓ 1
B1 angle ABC = (180° − 80°)/2 = 50°	✓ 1
A1 bearing 200.0°	✓ 1

Common error: The bearing of A from B is not the same as the bearing of B from A: add or subtract 180°.

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On a grid in a game, the player is at P(−3, 5) and a treasure chest is at C(9, −4), where each unit is 1 metre.

(a)

Calculate the distance PC.

(3)

(b)

Calculate the acute angle that the line PC makes with the horizontal, to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 horizontal difference 12 and vertical difference 9	✓ 1
M1 sqrt(12² + 9²)	✓ 1
A1 15 m	✓ 1
Part (b)
M1 tan x = 9/12	✓ 1
A1 36.9°	✓ 1

Common error: Subtract coordinates carefully: 9 − (−3) = 12, not 6.

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A water tower stands on level ground. From a point P, 40 m from the base of the tower and at the same level, the angle of elevation of the top of the tower is 32°.

Diagram not to scale

(a)

Calculate the height of the tower, to 3 significant figures.

(2)

(b)

Write down the angle of depression of P from the top of the tower.

(1)

(c)

A second observer stands in line with P and the base of the tower, on the same side of it as P. From there, the angle of elevation of the top is 51°. Calculate how far the second observer is from the base of the tower, to 3 significant figures. Use the unrounded height from (a).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 40 tan 32°	✓ 1
A1 25.0 m	✓ 1
Part (b)
B1 32° (alternate angles)	✓ 1
Part (c)
M1 d = height / tan 51°	✓ 1
A1 20.2 m	✓ 1

Exam tip: The angle of depression from the top equals the angle of elevation from the ground because they are alternate angles.

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A conveyor belt carries roasted coffee beans up a straight slope. The belt makes an angle of 28° with the horizontal and rises through a vertical height of 1.9 m. Give each answer correct to 3 significant figures.

Diagram not to scale

(a)

Calculate the length of the belt.

(2)

(b)

Calculate the horizontal distance covered by the belt.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sin 28° = 1.9 / L	✓ 1
A1 4.05 m	✓ 1
Part (b)
M1 tan 28° = 1.9 / x	✓ 1
A1 3.57 m	✓ 1

Common error: Set the calculator to degree mode before using sin, cos or tan with an angle in degrees.

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A survey drone flies 12 km from point A to point B on a bearing of 070°. It then flies 9 km from B to point C on a bearing of 160°.

Diagram not to scale

(a)

Show that angle ABC = 90°.

(2)

(b)

Find the distance AC.

(2)

(c)

Find the bearing of C from A, correct to the nearest degree.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 the bearing of A from B is 070° + 180° = 250°	✓ 1
B1 250° − 160° = 90°	✓ 1
Part (b)
M1 AC² = 12² + 9²	✓ 1
A1 15 km	✓ 1
Part (c)
M1 tan BAC = 9 / 12	✓ 1
A1 angle BAC = 36.9°	✓ 1
A1 bearing = 070° + 36.9° = 107°	✓ 1

Exam tip: Draw a north line at each point and mark every bearing clockwise from north before calculating.

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The top of a bridge tower is seen from a point on level ground 85 m from its base. The angle of elevation of the top of the tower is 31°.

Diagram not to scale

(a)

Calculate the height of the tower. Give your answer correct to 3 significant figures.

(2)

(b)

The observer walks directly towards the tower until the angle of elevation of the top is 50°. Calculate the new distance from the base of the tower, correct to 3 significant figures. Use the unrounded height from (a).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 height = 85 × tan 31°	✓ 1
A1 51.1 m	✓ 1
Part (b)
M1 tan 50° = 51.07 / x	✓ 1
M1 x = 51.07 / tan 50°	✓ 1
A1 42.9 m	✓ 1

Exam tip: Keep unrounded values from earlier parts in your calculator to avoid losing accuracy in the final answer.

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A bridge truss is an isosceles triangle with a base of 10 m and two equal sloping sides of 13 m.

Diagram not to scale

(a)

Show that the height of the truss is 12 m.

(2)

(b)

Calculate the area of the triangle.

(2)

(c)

Calculate the size of one of the base angles. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 height² = 13² − 5² = 144	✓ 1
A1 height = 12 m	✓ 1
Part (b)
M1 area = ½ × 10 × 12	✓ 1
A1 60 m²	✓ 1
Part (c)
M1 cos x = 5 / 13	✓ 1
A1 67.4°	✓ 1

Common error: The base of each right-angled triangle is half the base of the isosceles triangle, 5 m, not 10 m.

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A triangular brace in a bridge has sides of length 2.1 m, 2.8 m and 3.5 m.

Diagram not to scale

(a)

Show that the triangle is right-angled.

(2)

(b)

Calculate the smallest angle of the triangle. Give your answer correct to 1 decimal place.

(2)

(c)

Calculate the area of the triangle.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 2.1² + 2.8² = 4.41 + 7.84 = 12.25	✓ 1
A1 3.5² = 12.25, so by the converse of Pythagoras' theorem the triangle is right-angled	✓ 1
Part (b)
M1 sin x = 2.1 / 3.5	✓ 1
A1 36.9°	✓ 1
Part (c)
M1 area = ½ × 2.1 × 2.8	✓ 1
A1 2.94 m²	✓ 1

Common error: To test for a right angle, compare the sum of the two smaller squares with the square of the longest side.

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A rectangular solar panel of length 1.8 m is fixed on a sloping roof. The panel lies along the line of greatest slope of the roof, which makes an angle of 35° with the horizontal. Use unrounded values in later parts.

Diagram not to scale

(a)

Calculate the vertical height of the top edge of the panel above its bottom edge. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the horizontal distance between the top edge and the bottom edge of the panel. Give your answer correct to 3 significant figures.

(2)

(c)

The bottom edge of the panel is 2.5 m above the ground. Calculate the height of the top edge of the panel above the ground. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(1.8\sin35^\circ\)	✓ 1
A1 1.03 m	✓ 1
Part (b)
M1 \(1.8\cos35^\circ\)	✓ 1
A1 1.47 m	✓ 1
Part (c)
M1 \(2.5+1.8\sin35^\circ\) (unrounded value from (a))	✓ 1
A1 3.53 m	✓ 1

Common error: Using cos for the vertical height; the height is opposite the 35° angle, so it needs sin.

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A maintenance boat leaves a harbour A and sails 18 km on a bearing of 070° to a wind-farm platform B. It then sails 24 km on a bearing of 160° to a second platform C.

Diagram not to scale

(a)

Show that angle ABC = 90°.

(2)

(b)

Calculate the distance AC.

(2)

(c)

Calculate the bearing of C from A. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of A from B is \(70^\circ+180^\circ=250^\circ\)	✓ 1
A1 \(250^\circ-160^\circ=90^\circ\) shown	✓ 1
Part (b)
M1 \(\sqrt{18^2+24^2}\)	✓ 1
A1 30 km	✓ 1
Part (c)
M1 \(\tan^{-1}\left(\frac{24}{18}\right)\) for angle BAC	✓ 1
A1 angle BAC = 53.1°	✓ 1
A1 bearing = \(70^\circ+53.1^\circ=123.1^\circ\)	✓ 1

Common error: Subtracting 180° instead of adding it, or forgetting that the angle at B is found from the back bearing of A, not from 070° directly.

+ Add to test✓ In your test173IGMC3-6-09r19-06Pythagoras' theorem and right-angled trigonometry Paper CMedium6 marks

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A surveyor stands on level ground 40 m from the base of a wind turbine tower. The angle of elevation of the hub (the top of the tower) is 62°. The turbine has blades of length 38 m that rotate about the hub. Use unrounded values in later parts.

Diagram not to scale

(a)

Calculate the height of the hub above the ground. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the height above the ground of the tip of a blade when it points vertically downwards. Give your answer correct to 3 significant figures.

(2)

(c)

The surveyor now moves to a point on the ground 90 m from the base of the tower. Calculate the angle of elevation of the hub from this point. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(40\tan62^\circ\)	✓ 1
A1 75.2 m	✓ 1
Part (b)
M1 \(40\tan62^\circ-38\) (unrounded value from (a))	✓ 1
A1 37.2 m	✓ 1
Part (c)
M1 \(\tan^{-1}\left(\frac{75.23}{90}\right)\) (unrounded value from (a))	✓ 1
A1 39.9°	✓ 1

Exam tip: Keep unrounded values in your calculator for later parts; rounding too early can change the final digit.

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The floor of a shop is a rectangle ABCD with AB = 24 m and BC = 18 m. A point P lies on AB with AP = 16 m.

Diagram not to scale

(a)

Calculate the length of the diagonal AC.

(2)

(b)

Calculate angle CAB. Give your answer correct to 1 decimal place.

(2)

(c)

A display stand is placed at P and an electric cable is run from P to C. Calculate the length PC. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{24^2+18^2}\)	✓ 1
A1 30 m	✓ 1
Part (b)
M1 \(\tan^{-1}\left(\frac{18}{24}\right)\)	✓ 1
A1 36.9°	✓ 1
Part (c)
B1 \(PB=24-16=8\) m	✓ 1
M1 \(\sqrt{8^2+18^2}\) or \(\sqrt{388}\)	✓ 1
A1 19.7 m	✓ 1

Exam tip: When a point lies on a side of a rectangle, first work out the remaining length of that side before using Pythagoras' theorem.

+ Add to test✓ In your test175IGMC3-6-09r19-09Bearings Paper CMedium7 marks

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A drone delivers medical supplies from a base O. It flies 6.5 km on a bearing of 125° to a clinic D. Use unrounded values in later parts.

Diagram not to scale

(a)

Calculate how far east of O the clinic D is. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate how far south of O the clinic D is. Give your answer correct to 3 significant figures.

(2)

(c)

A second site H is 8.0 km due east of O. Calculate the distance DH. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(6.5\sin55^\circ\) (or \(6.5\sin125^\circ\))	✓ 1
A1 5.32 km	✓ 1
Part (b)
M1 \(6.5\cos55^\circ\) (or \(-6.5\cos125^\circ\))	✓ 1
A1 3.73 km	✓ 1
Part (c)
M1 east difference \(8-6.5\sin55^\circ=2.68\) and south difference \(3.73\) (unrounded values)	✓ 1
M1 \(\sqrt{(8-6.5\sin55^\circ)^2+(6.5\cos55^\circ)^2}\)	✓ 1
A1 4.59 km	✓ 1

Common error: Mixing up sin and cos: a bearing is measured from north, so the angle needed for the right-angled triangle is often 180° minus the bearing, not the bearing itself.

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An air ambulance flies from a base A on a bearing of 090° for 40 km to a village V. It then flies on a bearing of 180° for 30 km to a hospital H.

Diagram not to scale

(a)

Calculate the direct distance AH.

(2)

(b)

Show that the bearing of H from A is 126.9°, correct to 1 decimal place.

(3)

(c)

Find the bearing of A from H. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 40² + 30² = 2500	✓ 1
A1 50 km	✓ 1
Part (b)
M1 angle VAH, tan x = 30 ÷ 40	✓ 1
A1 x = 36.9° (36.87°)	✓ 1
A1 bearing = 90° + 36.9° = 126.9°	✓ 1
Part (c)
M1 126.9° + 180° (use of back bearing)	✓ 1
A1 306.9°	✓ 1

Exam tip: When a path turns through a right angle on a compass bearing, the triangle formed is right-angled, so Pythagoras and tan can be used directly.

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A camera is fixed at the top of a gantry 14 m above a flat straight pit lane. A race car stops on the pit lane in line with the base of the gantry. The angle of depression from the camera to the car is 22°. Use the unrounded value from part (a) in part (b).

Diagram not to scale

(a)

Calculate the horizontal distance from the base of the gantry to the car. Give your answer correct to 3 significant figures.

(2)

(b)

The car then drives towards the gantry until the angle of depression is 40°. Calculate how far the car has moved. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 tan 22° = 14 ÷ d	✓ 1
A1 34.7 m	✓ 1
Part (b)
M1 new distance = 14 ÷ tan 40°	✓ 1
M1 34.65 − 16.68	✓ 1
A1 18.0 m	✓ 1

Exam tip: The angle of depression is measured down from the horizontal, so it equals the angle of elevation from the car to the camera.

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A drone flies from the pit garage P due north for 300 m to a point A. It then flies 200 m on a bearing of 120° to a point B.

Diagram not to scale

(a)

Calculate how far east of A the point B is. Give your answer correct to 3 significant figures.

(2)

(b)

Show that B is 200 m north of P.

(2)

(c)

Calculate the distance PB. Use unrounded values and give your answer correct to 3 significant figures.

(2)

(d)

Calculate the bearing of B from P. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 200 sin 60° (or 200 sin 120°)	✓ 1
A1 173 m	✓ 1
Part (b)
M1 B is 200 cos 60° south of A, which is 100 m	✓ 1
A1 300 − 100 = 200 m	✓ 1
Part (c)
M1 PB = √(173.2…² + 200²)	✓ 1
A1 265 m	✓ 1
Part (d)
M1 tan x = 173.2… ÷ 200 (ECF)	✓ 1
A1 40.9°	✓ 1

Exam tip: Splitting each leg of a journey into east-west and north-south components lets you use right-angled triangles even when the path turns through an angle that is not 90°.

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A hospital wheelchair ramp rises at a constant angle of 5.2° to the horizontal. The ramp covers a horizontal distance of 15 m.

Diagram not to scale (vertical scale exaggerated)

(a)

Calculate the height gained by the ramp. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the length of the sloping surface of the ramp. Give your answer correct to 3 significant figures.

(2)

(c)

A guideline says that a ramp must not be steeper than 1 in 12 (a rise of 1 for every 12 along the horizontal). Show whether this ramp meets the guideline.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 height = 15 tan 5.2°	✓ 1
A1 1.37 m	✓ 1
Part (b)
M1 length = 15 ÷ cos 5.2°	✓ 1
A1 15.1 m	✓ 1
Part (c)
M1 tan 5.2° = 0.0910… compared with 1 ÷ 12 = 0.0833…	✓ 1
A1 0.0910… > 0.0833…, so the ramp is steeper than 1 in 12 and does not meet the guideline	✓ 1

Exam tip: To compare two slopes, compare their gradients (rise ÷ run) rather than the angles and lengths separately.

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A fashion student cuts an isosceles triangular panel for a scarf. The base is 48 cm and each of the two equal sides is 74 cm long.

Diagram not to scale

(a)

Show that the perpendicular height of the panel is 70 cm.

(3)

(b)

Calculate the area of the panel.

(2)

(c)

Calculate the size of one of the base angles. Give your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 split into two right-angled triangles with base 24 cm	✓ 1
M1 \(\sqrt{74^2-24^2}\)	✓ 1
A1 \(\sqrt{4900}=70\) cm	✓ 1
Part (b)
M1 \(\frac{1}{2}\times48\times70\)	✓ 1
A1 1680 cm²	✓ 1
Part (c)
M1 \(\cos^{-1}\left(\frac{24}{74}\right)\) (or \(\tan^{-1}\frac{70}{24}\))	✓ 1
A1 71.1°	✓ 1

Exam tip: In an 'show that' question, write the unrounded value (here an exact 70) before stating the result.

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A trawler leaves a harbour and sails 20 km on a bearing of 125°.

Diagram not to scale

(a)

The angle between the trawler's path and the line pointing due south is 55°. Calculate how far east the trawler is of the harbour, to 3 significant figures.

(2)

(b)

Calculate how far south the trawler is of the harbour, to 3 significant figures.

(2)

(c)

The trawler returns directly to the harbour. Write down the bearing on which it sails.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(20\sin55^\circ\)	✓ 1
A1 16.4 km	✓ 1
Part (b)
M1 \(20\cos55^\circ\)	✓ 1
A1 11.5 km	✓ 1
Part (c)
B1 305°	✓ 1

Exam tip: A back-bearing is found by adding or subtracting 180° so that the result stays between 000° and 360°.

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A maintenance robot climbs a straight ramp of length 3.5 m. The top of the ramp is 0.84 m above the ground.

Diagram not to scale

(a)

Calculate the angle the ramp makes with the horizontal, to 1 decimal place.

(2)

(b)

Calculate the horizontal distance covered by the ramp, to 3 significant figures.

(2)

(c)

The robot cannot climb a slope steeper than 10° to the horizontal. A new ramp of the same length 3.5 m is built at exactly 10°. Calculate its height, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sin^{-1}\left(\frac{0.84}{3.5}\right)\)	✓ 1
A1 13.9°	✓ 1
Part (b)
M1 \(\sqrt{3.5^2-0.84^2}\) (or \(3.5\cos13.89^\circ\))	✓ 1
A1 3.40 m	✓ 1
Part (c)
M1 \(3.5\sin10^\circ\)	✓ 1
A1 0.608 m	✓ 1

Common error: Using \(\cos\) instead of \(\sin\) for the height: the height is opposite the angle, so it goes with sine when the ramp is the hypotenuse.

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A helicopter flies from an airfield A on a bearing of 070° for 80 km to a farm B. It then flies on a bearing of 160° for 60 km to a village C.

Diagram not to scale

(a)

Show that angle ABC = 90°.

(2)

(b)

Calculate the straight-line distance AC.

(2)

(c)

Calculate the bearing of C from A. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of A from B is 250° (070 + 180) or co-interior angles used	✓ 1
A1 250 − 160 = 90°	✓ 1
Part (b)
M1 sqrt(80² + 60²)	✓ 1
A1 100 km	✓ 1
Part (c)
M1 angle BAC from tan = 60/80	✓ 1
A1 BAC = 36.9°	✓ 1
B1 bearing = 070° + 36.9° = 106.9° (accept 106.87...)	✓ 1

Exam tip: Draw a north line at every point and mark the angles on the sketch before calculating.

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At a music festival a vertical pole 7.2 m tall stands on level ground and supports a speaker. A straight support wire runs from the top of the pole to a point on the ground 9.6 m from the base of the pole.

Diagram not to scale

(a)

Calculate the length of the wire.

(2)

(b)

Calculate the angle the wire makes with the ground. Give your answer correct to 1 decimal place.

(2)

(c)

A second wire runs from the top of the pole to the ground on the opposite side of the pole and makes an angle of 50° with the ground. Calculate the distance of its ground end from the base of the pole. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sqrt(7.2² + 9.6²)	✓ 1
A1 12 m	✓ 1
Part (b)
M1 tan x = 7.2/9.6 (or sin x = 7.2/12)	✓ 1
A1 36.9°	✓ 1
Part (c)
M1 tan 50 = 7.2/d	✓ 1
M1 d = 7.2 ÷ tan 50	✓ 1
A1 6.04 m	✓ 1

Exam tip: In a right-angled triangle label the sides opposite, adjacent and hypotenuse relative to the angle used.

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A farmer leaves a gate G and walks 450 m on a bearing of 054° to a well W. From W she walks 300 m due south to a barn P. Use unrounded values in your working and round only your final answers.

Diagram not to scale

(a)

Calculate how far east of G the well W is. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate how far north of G the well W is. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the bearing of P from G. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 450 × sin 54°	✓ 1
A1 364 m	✓ 1
Part (b)
M1 450 × cos 54°	✓ 1
A1 265 m	✓ 1
Part (c)
B1 P is 450cos54° − 300 = −35.5 m north of G, so 35.5 m south	✓ 1
M1 angle below the east direction = tan⁻¹(35.5/364.06) = 5.6°	✓ 1
A1 bearing 095.6°	✓ 1

Common error: Resolve each leg into east and north components, then combine them.

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A drone surveys a vineyard. It flies from A 800 m due north to B, then 600 m due west to C.

Diagram not to scale

(a)

Calculate the distance AC.

(2)

(b)

Calculate the bearing of C from A. Give your answer correct to 1 decimal place.

(3)

(c)

Write down the bearing of A from C, correct to 1 decimal place.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sqrt(800² + 600²)	✓ 1
A1 1000 m	✓ 1
Part (b)
M1 tan x = 600/800 for the angle at A	✓ 1
A1 x = 36.9°	✓ 1
A1 bearing = 360 − 36.9 = 323.1°	✓ 1
Part (c)
B1 143.1° (ECF from (b))	✓ 1

Common error: A back bearing differs from the forward bearing by 180°: subtract if the forward bearing is more than 180°.

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A light aircraft takes off and climbs in a straight line at 14° above the horizontal. It flies 5.5 km along this straight path.

Diagram not to scale

(a)

Calculate the height it has gained. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the horizontal distance it has covered. Give your answer correct to 3 significant figures.

(2)

(c)

The aircraft's speed along its path is 220 km/h. Calculate the time taken to fly the 5.5 km. Give your answer in minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 5.5 × sin 14°	✓ 1
A1 1.33 km	✓ 1
Part (b)
M1 5.5 × cos 14°	✓ 1
A1 5.34 km	✓ 1
Part (c)
M1 5.5 ÷ 220 × 60	✓ 1
A1 1.5 minutes	✓ 1

Exam tip: Check whether the distance you are given is along the path or horizontal before choosing sin or cos.

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A lorry leaves a depot and drives 45 km on a bearing of 090°. It then drives 60 km on a bearing of 180°.

Diagram not to scale

(a)

Calculate the distance of the lorry from the depot.

(2)

(b)

Find the bearing of the depot from the lorry, to the nearest degree.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{45^2 + 60^2}\)	✓ 1
A1 75 km	✓ 1
Part (b)
M1 \(\tan^{-1}\left(\frac{45}{60}\right)\) or an equivalent angle \(= 36.9°\)	✓ 1
M1 \(360° - 36.9°\) (north-west direction)	✓ 1
A1 323°	✓ 1

Exam tip: Draw a quick sketch with a north line at each turning point; the final bearing is measured clockwise from north, so the answer is rarely the angle in the triangle.

+ Add to test✓ In your test189IGMC3-6-10r25-05Pythagoras' theorem and right-angled trigonometry Paper CMedium6 marks

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The end wall of a building is an isosceles triangle with a horizontal base of 9.6 m. The two equal sloping sides are each 7.4 m long.

Diagram not to scale

(a)

Calculate the perpendicular height of the wall. Give your answer to 3 significant figures.

(2)

(b)

Calculate the area of the wall. Use the unrounded value of the height from (a). Give your answer to 3 significant figures.

(2)

(c)

Calculate the angle between a sloping side and the base. Give your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{7.4^2 - 4.8^2}\)	✓ 1
A1 5.63 m	✓ 1
Part (b)
M1 \(\frac{1}{2} \times 9.6 \times 5.632\)	✓ 1
A1 27.0 m²	✓ 1
Part (c)
M1 \(\cos^{-1}\left(\frac{4.8}{7.4}\right)\)	✓ 1
A1 49.6°	✓ 1

Common error: Use half the base (4.8 m) in the right-angled triangle, not the whole base.

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An architect stands 40 m from the base of a tower on level ground. Her eyes are 1.6 m above the ground. The angle of elevation of the top of the tower from her eyes is 52°.

Diagram not to scale

(a)

Calculate the height of the top of the tower above her eye level. Give your answer to 3 significant figures.

(2)

(b)

Find the height of the tower above the ground.

(1)

(c)

She walks away from the tower along level ground until the angle of elevation of the top is 30°. Calculate her new distance from the base of the tower. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(40 \tan 52°\)	✓ 1
A1 51.2 m	✓ 1
Part (b)
B1 \(51.2 + 1.6 = 52.8\) m	✓ 1
Part (c)
M1 uses the height above eye level \(51.2\) in \(\tan 30° = \frac{51.2}{d}\)	✓ 1
M1 \(d = \frac{51.2}{\tan 30°}\)	✓ 1
A1 88.7 m	✓ 1

Common error: The angle of elevation is measured from her eyes, so the 1.6 m must be added only after finding the height above eye level.

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Port \(Q\) is 18 km due south of port \(R\). A cargo ship \(S\) is 24 km due east of \(Q\).

Diagram not to scale

(a)

Calculate the distance \(RS\).

(2)

(b)

Find the bearing of \(S\) from \(R\), to the nearest degree.

(3)

(c)

The ship sails directly from \(S\) to \(R\) at a constant speed of 15 km/h. Find the time taken.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{18^2 + 24^2}\)	✓ 1
A1 30 km	✓ 1
Part (b)
M1 \(\tan^{-1}\left(\frac{24}{18}\right) = 53.1°\) for angle \(QRS\)	✓ 1
M1 \(180° - 53.1°\)	✓ 1
A1 127°	✓ 1
Part (c)
M1 \(\frac{30}{15}\) (ECF from (a))	✓ 1
A1 2 hours	✓ 1

Exam tip: A bearing is measured clockwise from north, so with \(S\) to the south-east of \(R\) the answer lies between 090° and 180°.

+ Add to test✓ In your test192IGMC3-6-08r26-03Bearings Paper CMedium8 marks

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A ship carrying coffee beans sails from port P on a bearing of 065° for 52 km to a buoy Q. It then sails on a bearing of 155° for 31 km to a port R.

Diagram not to scale

(a)

Show that angle PQR = 90°.

(2)

(b)

Calculate the distance PR. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the bearing of R from P. Give your answer correct to 1 decimal place.

(3)

(d)

Write down the bearing of P from R.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 the bearing of P from Q is 065° + 180° = 245°	✓ 1
A1 245° − 155° = 90°	✓ 1
Part (b)
M1 52² + 31² = 3665	✓ 1
A1 60.5 km	✓ 1
Part (c)
M1 tan(angle RPQ) = 31 ÷ 52	✓ 1
A1 angle RPQ = 30.8°	✓ 1
A1 065° + 30.8° = 095.8°	✓ 1
Part (d)
B1 275.8° (follow through from (c): bearing in (c) + 180°)	✓ 1

Exam tip: Bearings are measured clockwise from north and are written with three figures. Draw a north line at each point before you work out an angle.

+ Add to test✓ In your test193IGMC3-6-08r26-05Pythagoras' theorem and right-angled trigonometry Paper CMedium6 marks

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A bridge pylon stands in the water with its top 38 m above the level of the water. Boat A is on one side of the pylon and boat B is on the other side. The base of the pylon is on the straight line AB, between the two boats. From the top of the pylon, the angle of depression of boat A is 28° and the angle of depression of boat B is 17°.

Diagram not to scale

(a)

Calculate the horizontal distance from boat A to the base of the pylon. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate the distance AB. Give your answer correct to the nearest metre.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 the angle of elevation of the top of the pylon from A is 28° (alternate angles)	✓ 1
M1 38 ÷ tan 28°	✓ 1
A1 71.5 m	✓ 1
Part (b)
M1 distance from B to the base = 38 ÷ tan 17° (= 124.3)	✓ 1
M1 adding the two distances, 71.5 + 124.3	✓ 1
A1 196 m	✓ 1

Common error: The angle of depression is measured from the horizontal at the top, so it equals the angle of elevation at the boat; do not use 90° minus this angle.

+ Add to test✓ In your test194IGMC3-6-08r26-06Pythagoras' theorem and right-angled trigonometry Paper CMedium6 marks

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The roof of a coffee roastery is supported by a frame in the shape of an isosceles triangle. The base of the frame is 14 m and each of the two equal sloping sides is 9 m.

Diagram not to scale

(a)

Calculate the angle between a sloping side and the base. Give your answer correct to 1 decimal place.

(2)

(b)

Calculate the vertical height of the frame. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the area of the triangle. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cos x = 7 ÷ 9 (using half of the base)	✓ 1
A1 38.9°	✓ 1
Part (b)
M1 √(9² − 7²) = √32	✓ 1
A1 5.66 m	✓ 1
Part (c)
M1 ½ × 14 × 5.66 (use of unrounded height also accepted)	✓ 1
A1 39.6 m²	✓ 1

Exam tip: Drop a perpendicular from the top of an isosceles triangle to split it into two identical right-angled triangles.

+ Add to test✓ In your test195IGMC3-6-08r26-14Pythagoras' theorem and right-angled trigonometry Paper CMedium7 marks

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A ramp is used to load sacks of coffee onto a lorry. The ramp is a straight slope 4.2 m long, and its top end is 0.95 m above the ground.

Diagram not to scale

(a)

Calculate the angle that the ramp makes with the ground. Give your answer correct to 1 decimal place.

(2)

(b)

Calculate the horizontal distance from the bottom of the ramp to the point on the ground directly below the top. Give your answer correct to 3 significant figures.

(2)

(c)

A safety rule says that the ramp must make an angle of no more than 12° with the ground. State whether this ramp meets the rule, giving a reason.

(1)

(d)

A longer ramp is used with its top end still 0.95 m above the ground and an angle of exactly 12°. Calculate the length of this ramp. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sin x = 0.95 ÷ 4.2	✓ 1
A1 13.1°	✓ 1
Part (b)
M1 √(4.2² − 0.95²)	✓ 1
A1 4.09 m	✓ 1
Part (c)
B1 no, because 13.1° is greater than 12° (follow through from (a))	✓ 1
Part (d)
M1 length = 0.95 ÷ sin 12°	✓ 1
A1 4.57 m	✓ 1

Exam tip: When the unknown length is in the denominator, rearrange the sine ratio first: length = opposite ÷ sin x.

+ Add to test✓ In your test196IGMC3-6-11r27-05Sine rule, cosine rule, trigonometric graphs and 3D problems Paper CMedium6 marks

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A triangular field ABC is isosceles with AB = AC = 100 m and BC = 120 m. The midpoint of BC is M.

Diagram not to scale

(a)

Calculate the perpendicular height AM of the field.

(2)

(b)

Calculate angle ABC. Give your answer to 1 decimal place.

(2)

(c)

Seed costs €0.85 for every 100 m² of field. Calculate the cost of seed for the whole field.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 AM² = 100² − 60²	✓ 1
A1 80 m	✓ 1
Part (b)
M1 cos B = 60 ÷ 100 (or sin B = 80 ÷ 100, or tan B = 80 ÷ 60)	✓ 1
A1 53.1°	✓ 1
Part (c)
M1 area = ½ × 120 × 80 (= 4800 m²) and 4800 ÷ 100 × 0.85	✓ 1
A1 €40.80	✓ 1

Common error: The height of an isosceles triangle splits the base in half: use 60 m, not 120 m, in the right-angled triangle.

+ Add to test✓ In your test197IGMC3-6-11r27-06Bearings Paper CMedium6 marks

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A crop-spraying drone takes off from a point O and flies in a straight line for 3.2 km on a bearing of 130° to a point A.

Diagram not to scale

(a)

Calculate how far east of O the point A is.

(2)

(b)

Calculate how far south of O the point A is.

(2)

(c)

Write down the bearing of O from A.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 3.2 × sin 130° (or 3.2 × cos 40°)	✓ 1
A1 2.45 km	✓ 1
Part (b)
M1 3.2 × cos 50° (or 3.2 × sin 40°, or −3.2 × cos 130°)	✓ 1
A1 2.06 km	✓ 1
Part (c)
M1 130° + 180°	✓ 1
A1 310°	✓ 1

Exam tip: To find a back bearing, add 180° if the bearing is less than 180°, and subtract 180° if it is more.

+ Add to test✓ In your test198IGMC3-6-11r27-07Pythagoras' theorem and right-angled trigonometry Paper CMedium6 marks

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A ramp is used to load bales of hay onto a trailer. The ramp is a straight plank of length 4.5 m and it rises through a vertical height of 1.2 m.

Diagram not to scale

(a)

Calculate the angle between the ramp and the horizontal. Give your answer to 1 decimal place.

(2)

(b)

For safety, the ramp must make an angle of 12° with the horizontal. The height stays at 1.2 m. Calculate the length of the new plank. Give your answer to 3 significant figures.

(2)

(c)

Calculate the horizontal distance from the foot of the new ramp to the point directly below its top end. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sin⁻¹(1.2 ÷ 4.5)	✓ 1
A1 15.5°	✓ 1
Part (b)
M1 length = 1.2 ÷ sin 12°	✓ 1
A1 5.77 m	✓ 1
Part (c)
M1 distance = 1.2 ÷ tan 12° (or 5.77… × cos 12°)	✓ 1
A1 5.65 m	✓ 1

Common error: Keep unrounded values in your calculator for later parts; 5.77 × cos 12° gives a slightly different last digit if you use the rounded plank length.

+ Add to test✓ In your test199IGMC3-6-11r27-08Sine rule, cosine rule, trigonometric graphs and 3D problems Paper CMedium7 marks

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The stage in a concert hall is the shape of an isosceles trapezium ABCD. The front edge AB is parallel to the back edge DC. AB = 14 m, DC = 8 m and the sloping sides AD = BC = 5 m.

Diagram not to scale

(a)

Show that the perpendicular distance between AB and DC is 4 m.

(3)

(b)

Calculate angle DAB. Give your answer to 1 decimal place.

(2)

(c)

Calculate the area of the stage.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 horizontal overhang at each end = (14 − 8) ÷ 2 = 3 m	✓ 1
M1 height² = 5² − 3²	✓ 1
A1 height = sqrt(16) = 4 m	✓ 1
Part (b)
M1 cos A = 3 ÷ 5 (or tan A = 4 ÷ 3, or sin A = 4 ÷ 5)	✓ 1
A1 53.1°	✓ 1
Part (c)
M1 ½ × (14 + 8) × 4	✓ 1
A1 44 m²	✓ 1

Exam tip: In a symmetric trapezium, drop a perpendicular from each end of the shorter side to make two identical right-angled triangles and a rectangle.

+ Add to test✓ In your test200IGMC3-6-11r27-13Pythagoras' theorem and right-angled trigonometry Paper CMedium6 marks

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The top of a wind turbine on a farm is seen from a point P on level ground. P is 80 m from the foot of the turbine, and the angle of elevation of the top is 28°.

Diagram not to scale

(a)

Calculate the height of the turbine. Give your answer to 3 significant figures.

(2)

(b)

A second point Q is in line with P and the turbine, further from the turbine. The angle of elevation of the top from Q is 15°. Use the unrounded height to calculate the distance from Q to the foot of the turbine. Give your answer to 3 significant figures.

(2)

(c)

Calculate the distance PQ. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 height = 80 × tan 28°	✓ 1
A1 42.5 m	✓ 1
Part (b)
M1 distance = 42.53… ÷ tan 15°	✓ 1
A1 159 m	✓ 1
Part (c)
M1 158.7… − 80	✓ 1
A1 78.7 m	✓ 1

Common error: Subtract the unrounded distance (158.7…) rather than 159, otherwise PQ becomes 79 m instead of 78.7 m.

+ Add to test✓ In your test201IGMC3-6-10r30-03Bearings Paper CMedium5 marks

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A tracking telescope at point B is 8.0 km from a telescope at point A, on a bearing of 040° from A.

(a)

Calculate how far B is north of A and how far B is east of A. Give each answer correct to 3 significant figures.

(4)

(b)

Write down the bearing of A from B.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 north = \(8.0\cos 40°\)	✓ 1
A1 6.13 km	✓ 1
M1 east = \(8.0\sin 40°\)	✓ 1
A1 5.14 km	✓ 1
Part (b)
B1 220°	✓ 1

Exam tip: A bearing is measured clockwise from north, so the north component uses cosine and the east component uses sine.

+ Add to test✓ In your test202IGMC3-6-10r30-04Bearings Paper CMedium6 marks

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An inspection boat leaves a point P beneath a bridge and sails 5.0 km due east and then 12 km due north to a point Q.

(a)

Calculate the distance PQ.

(2)

(b)

Calculate the bearing of Q from P. Give your answer correct to 1 decimal place.

(3)

(c)

Calculate the bearing of P from Q. Give your answer correct to 1 decimal place.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{5^2+12^2}\)	✓ 1
A1 13 km	✓ 1
Part (b)
M1 angle from north \(=\tan^{-1}\left(\frac{5}{12}\right)\)	✓ 1
A1 22.6°	✓ 1
A1 bearing 022.6°	✓ 1
Part (c)
B1 202.6° (ECF from (b))	✓ 1

Common error: A bearing is measured clockwise from north, so give three figures and add 180° to find the reverse bearing when the first is under 180°.

+ Add to test✓ In your test203IGMC3-6-10r30-06Pythagoras' theorem and right-angled trigonometry Paper CMedium5 marks

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A point on level ground is 40 m from the foot of a vertical bridge tower. The angle of elevation of the top of the tower from this point is 52°. A boat is on the river on the opposite side of the tower from the point, and the angle of depression of the boat from the top of the tower is 23°. The ground and the river are at the same level.

(a)

Calculate the height of the tower. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the horizontal distance of the boat from the foot of the tower. Use the unrounded height from part (a) and give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(40\tan 52°\)	✓ 1
A1 51.2 m	✓ 1
Part (b)
M1 angle of elevation of top from boat is 23° (alternate angles)	✓ 1
M1 \(\frac{51.1977...}{\tan 23°}\)	✓ 1
A1 121 m	✓ 1

Common error: The angle of depression is measured from the horizontal at the top, not from the vertical; it equals the angle of elevation from the boat.

+ Add to test✓ In your test204IGMC3-6-10r30-07Sine rule, cosine rule, trigonometric graphs and 3D problems Paper CMedium6 marks

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The reflector of a space probe antenna is supported by a triangular frame ABC with AB = AC = 10 cm and BC = 12 cm.

(a)

Calculate the perpendicular height of A above BC.

(2)

(b)

Calculate angle ABC. Give your answer correct to 1 decimal place.

(2)

(c)

Calculate the area of the triangle ABC.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 midpoint of BC, \(\sqrt{10^2-6^2}\)	✓ 1
A1 8.00 cm	✓ 1
Part (b)
M1 \(\cos ABC=\frac{6}{10}\)	✓ 1
A1 53.1°	✓ 1
Part (c)
M1 \(\frac{1}{2}\times 12\times 8\)	✓ 1
A1 48.0 cm² (ECF from (a))	✓ 1

Exam tip: Splitting an isosceles triangle down its line of symmetry gives two identical right-angled triangles.

+ Add to test✓ In your test205IGMC3-6-11r33-03Pythagoras' theorem and right-angled trigonometry Paper CMedium6 marks

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A water tower stands on level ground. Its top T is 18 m above the ground, vertically above the point F at its base. A straight pipe runs from T to a point P on the ground, with FP = 24 m.

(a)

Calculate the length of the pipe TP.

(2)

(b)

Calculate the angle of depression of P from T. Give your answer correct to 1 decimal place.

(2)

(c)

A point Q is on the ground on the same side of the tower as P. The angle of elevation of T from Q is 25°. Calculate FQ, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(18² + 24²)	✓ 1
A1 30 m	✓ 1
Part (b)
M1 angle TPF = tan⁻¹(18/24) (alternate angles equal)	✓ 1
A1 36.9°	✓ 1
Part (c)
M1 tan 25° = 18/FQ	✓ 1
A1 FQ = 18/tan 25°, 38.6 m	✓ 1

Exam tip: The angle of depression from the top equals the angle of elevation from the ground point because they are alternate angles.

+ Add to test✓ In your test206IGMC3-6-11r33-04Bearings Paper CMedium7 marks

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A survey team leaves a depot D and walks 8 km on a bearing of 040° to a pumping station S. It then walks 6 km on a bearing of 130° to a borehole B.

(a)

Show that angle DSB = 90°.

(2)

(b)

Calculate the distance DB.

(2)

(c)

Calculate the bearing of B from D. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of D from S is 040° + 180° = 220°	✓ 1
A1 220° − 130° = 90°	✓ 1
Part (b)
M1 √(8² + 6²)	✓ 1
A1 10 km	✓ 1
Part (c)
M1 angle SDB = tan⁻¹(6/8) = 36.87...°	✓ 1
M1 040° + 36.87...°	✓ 1
A1 076.9°	✓ 1

Common error: To find a back bearing, add or subtract 180° and check the result lies between 000° and 360°.

+ Add to test✓ In your test207IGMC3-6-11r33-10Pythagoras' theorem and right-angled trigonometry Paper CMedium6 marks

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A pharmacy sign is fixed to a vertical wall. The point P is on level ground 12 m from the foot of the wall. From P, the angle of elevation of the bottom of the sign is 9° and the angle of elevation of the top of the sign is 28°.

(a)

Calculate the height of the top of the sign above the ground. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the height of the bottom of the sign above the ground. Give your answer correct to 3 significant figures.

(2)

(c)

Hence find the height of the sign, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12 tan 28°	✓ 1
A1 6.38 m	✓ 1
Part (b)
M1 12 tan 9°	✓ 1
A1 1.90 m	✓ 1
Part (c)
M1 6.38... − 1.90...	✓ 1
A1 4.48 m	✓ 1

Common error: Do not use the difference of the two angles to find the height of the sign; subtract the two heights instead.

+ Add to test✓ In your test208IGMC3-6-11r33-11Bearings Paper CMedium6 marks

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A water well W and a storage tank T are on level ground. The bearing of T from W is 238° and WT = 450 m.

(a)

Write down the bearing of W from T.

(1)

(b)

Calculate how far T is due south of W. Give your answer correct to 3 significant figures.

(3)

(c)

Calculate how far T is due west of W. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 058°	✓ 1
Part (b)
M1 angle between WT and the south direction = 238° − 180° = 58°	✓ 1
M1 450 cos 58°	✓ 1
A1 238 m	✓ 1
Part (c)
M1 450 sin 58°	✓ 1
A1 382 m	✓ 1

Exam tip: Sketch the north–south line through the start point and mark the angle from it to the path; cos gives the component along that line.

+ Add to test✓ In your test209IGMC3-6-10r38-05Bearings Paper CMedium7 marks

Calculator

A research boat sails from a harbour H on a bearing of 040° for 12 km to a reef A. It then sails 9 km on a bearing of 130° to a feeding ground B.

Diagram not to scale

(a)

Show that angle HAB = 90°.

(2)

(b)

Calculate the direct distance HB.

(2)

(c)

Calculate the bearing of B from H, giving your answer to the nearest degree.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of H from A = 040° + 180° = 220° (co-interior angles or back bearing)	✓ 1
A1 220° − 130° = 90°, shown	✓ 1
Part (b)
M1 12² + 9² (= 225)	✓ 1
A1 15 km	✓ 1
Part (c)
M1 angle AHB = tan⁻¹(9 ÷ 12) (= 36.87...°)	✓ 1
M1 040° + angle AHB	✓ 1
A1 77° (077°)	✓ 1

Exam tip: Draw a north line at each turning point; the angle between a bearing and its back bearing is always 180°.

+ Add to test✓ In your test210IGMC3-6-10r38-06Pythagoras' theorem and right-angled trigonometry Paper CMedium7 marks

Calculator

From a point P on level ground, 40 m from the base of a tower, the angle of elevation of the top of the tower from eye level is 52°. The observer's eyes are 1.6 m above the ground. Give answers to 3 significant figures.

Diagram not to scale

(a)

Calculate the height of the top of the tower above the observer's eye level.

(2)

(b)

Calculate the height of the tower above the ground.

(2)

(c)

The observer walks directly towards the tower until the angle of elevation of the top is 70°. Use the unrounded height above eye level from part (a) to calculate how far the observer has walked.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 40 × tan 52°	✓ 1
A1 51.2 m (51.20... before rounding)	✓ 1
Part (b)
M1 height above eye level + 1.6	✓ 1
A1 52.8 m	✓ 1
Part (c)
M1 horizontal distance from tower = 51.20... ÷ tan 70° (= 18.63...)	✓ 1
M1 40 − horizontal distance	✓ 1
A1 21.4 m	✓ 1

Common error: The angle of elevation is measured from the observer's eye, so remember to add the eye height to reach the height above ground.

+ Add to test✓ In your test211IGMC3-6-10r38-07Pythagoras' theorem and right-angled trigonometry Paper CMedium5 marks

Calculator

In a distribution centre a straight conveyor belt carries parcels up to a platform. The belt makes an angle of 18° with the horizontal. Give answers to 3 significant figures.

Diagram not to scale

(a)

The belt is 9.5 m long. Calculate the height it raises the parcels.

(2)

(b)

A second belt at the same angle must raise parcels to a platform 3.5 m high. Calculate the length of this belt.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 9.5 × sin 18°	✓ 1
A1 2.94 m	✓ 1
Part (b)
M1 sin 18° = 3.5 ÷ L	✓ 1
M1 L = 3.5 ÷ sin 18°	✓ 1
A1 11.3 m	✓ 1

Common error: When the unknown is the hypotenuse, divide by sin θ; multiplying by sin θ gives a length that is too small.

+ Add to test✓ In your test212IGMC3-6-10r38-14Pythagoras' theorem and right-angled trigonometry Paper CMedium6 marks

Calculator

A roof truss is an isosceles triangle. The two equal sides are each 6.5 m long and the two equal angles at the base are each 32°. Give answers to 3 significant figures.

Diagram not to scale

(a)

Calculate the perpendicular height of the truss.

(2)

(b)

Calculate the length of the base of the truss.

(2)

(c)

Use the unrounded values from parts (a) and (b) to calculate the area of the triangle.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 6.5 × sin 32°	✓ 1
A1 3.44 m (3.444... before rounding)	✓ 1
Part (b)
M1 2 × 6.5 × cos 32°	✓ 1
A1 11.0 m (11.025... before rounding)	✓ 1
Part (c)
M1 ½ × base × height	✓ 1
A1 19.0 m²	✓ 1

Exam tip: Drop a perpendicular from the apex to split the isosceles triangle into two identical right-angled triangles.

+ Add to test✓ In your test213IGMC3-6-10r40-03Bearings Paper CMedium5 marks

Calculator

Camp B is 8.0 km from Camp A on a bearing of 065°.

Diagram not to scale

(a)

Write down the bearing of Camp A from Camp B.

(1)

(b)

Calculate how far east of Camp A Camp B is. Give your answer to 3 significant figures.

(2)

(c)

Calculate how far north of Camp A Camp B is. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 245°	✓ 1
Part (b)
M1 \(8\sin 65°\)	✓ 1
A1 7.25 km	✓ 1
Part (c)
M1 \(8\cos 65°\)	✓ 1
A1 3.38 km	✓ 1

Common error: Bearings are measured clockwise from north, so the northward distance uses cosine and the eastward distance uses sine.

+ Add to test✓ In your test214IGMC3-6-10r40-04Pythagoras' theorem and right-angled trigonometry Paper CMedium5 marks

Calculator

Three telecoms towers P, Q and R are on level ground. Tower Q is 12 km due north of tower P. Tower R is 16 km due east of tower Q.

Diagram not to scale

(a)

Calculate the distance PR.

(2)

(b)

Calculate the bearing of R from P. Give your answer to 1 decimal place.

(2)

(c)

Find the bearing of P from R. Give your answer to 1 decimal place.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{12^2+16^2}\)	✓ 1
A1 20 km	✓ 1
Part (b)
M1 \(\tan x=\dfrac{16}{12}\) where \(x\) is the angle at P between PQ and PR	✓ 1
A1 053.1°	✓ 1
Part (c)
B1 233.1° (accept 180° + their answer to (b))	✓ 1

Exam tip: Bearings are written with three figures, so 53.1° must be written as 053.1°.

+ Add to test✓ In your test215IGMC3-6-10r40-06Pythagoras' theorem and right-angled trigonometry Paper CMedium5 marks

Calculator

A climber stands on level ground 40 m from the foot of a vertical cliff. The angle of elevation of the top of the cliff from the climber is 52°.

Diagram not to scale

(a)

Calculate the height of the cliff. Give your answer to 3 significant figures.

(2)

(b)

The climber walks directly away from the cliff until the angle of elevation of the top is 40°. Calculate how far the climber walks. Give your answer to 3 significant figures. Use the unrounded height from part (a) in your working.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(40\tan 52°\)	✓ 1
A1 51.2 m	✓ 1
Part (b)
M1 new distance from cliff \(=\dfrac{40\tan 52°}{\tan 40°}\)	✓ 1
M1 subtract 40 from the new distance	✓ 1
A1 21.0 m	✓ 1

Exam tip: Keep the unrounded height on your calculator between parts so that rounding errors do not build up.

+ Add to test✓ In your test216IGMC3-6-10r40-09Pythagoras' theorem and right-angled trigonometry Paper CMedium4 marks

Calculator

A climber stands at the top of a vertical cliff that is 55 m high. The ground at the foot of the cliff is level.

Diagram not to scale

(a)

A tent is on the ground 90 m from the foot of the cliff. Calculate the angle of depression of the tent from the climber. Give your answer to 1 decimal place.

(2)

(b)

A van is parked on the same level ground so that the angle of depression of the van from the climber is 24°. Calculate the distance of the van from the foot of the cliff. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\tan x=\dfrac{55}{90}\)	✓ 1
A1 31.4°	✓ 1
Part (b)
M1 \(\dfrac{55}{\tan 24°}\)	✓ 1
A1 124 m	✓ 1

Common error: The angle of depression is measured down from the horizontal, so it equals the angle of elevation from the ground and not the angle at the top between the cliff face and the line of sight.

+ Add to test✓ In your test217IGMC3-6-11r40-03Pythagoras' theorem and right-angled trigonometry Paper CMedium4 marks

Calculator

A coffee roaster has a conical hopper. A vertical cross-section through the apex is an isosceles triangle. The two equal sloping sides each have length 40 cm and the width of the base is 30 cm.

Vertical cross-section through the apex. Diagram not to scale

(a)

Calculate the vertical height of the hopper. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the angle between a sloping side and the vertical axis of the hopper. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{40^2-15^2}\) or \(\sqrt{1375}\)	✓ 1
A1 37.1 cm	✓ 1
Part (b)
M1 \(\sin\theta=\frac{15}{40}\) or \(\tan\theta=\frac{15}{37.08}\) (accept use of 37.1)	✓ 1
A1 22.0°	✓ 1

Common error: Using the whole base width of 30 cm instead of half of it. In an isosceles triangle the height splits the base into two equal halves.

+ Add to test✓ In your test218IGMC3-6-11r40-04Pythagoras' theorem and right-angled trigonometry Paper CMedium5 marks

Calculator

A footbridge truss is a triangle with sides of length 7.5 m, 10 m and 12.5 m.

Diagram not to scale

(a)

Show that the triangle is right-angled.

(3)

(b)

Calculate the size of the smallest angle in the triangle. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(7.5^2+10^2=156.25\)	✓ 1
A1 \(12.5^2=156.25\)	✓ 1
R1 sum of squares of the two shorter sides equals the square of the longest side, so by the converse of Pythagoras' theorem the triangle is right-angled	✓ 1
Part (b)
M1 \(\sin\theta=\frac{7.5}{12.5}\) or \(\tan\theta=\frac{7.5}{10}\) or \(\cos\theta=\frac{10}{12.5}\)	✓ 1
A1 36.9°	✓ 1

Exam tip: In a "show that" question write out both sides of the comparison fully and finish with a short conclusion in words.

+ Add to test✓ In your test219IGMC3-6-11r40-05Bearings Paper CMedium6 marks

Calculator

A ship sails from port P for 24 km due east to a point Q. It then sails 10 km due south to a point R.

Diagram not to scale

(a)

Calculate the distance PR.

(2)

(b)

Calculate angle QPR. Give your answer correct to 1 decimal place.

(2)

(c)

Find the bearing of R from P. Use the value of angle QPR from part (b).

(1)

(d)

Find the bearing of P from R. Give your answer correct to 1 decimal place.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{24^2+10^2}\)	✓ 1
A1 26 km	✓ 1
Part (b)
M1 \(\tan(\text{QPR})=\frac{10}{24}\)	✓ 1
A1 22.6°	✓ 1
Part (c)
B1 90° + 22.6° = 112.6° (ECF from (b))	✓ 1
Part (d)
B1 112.6° + 180° = 292.6° (ECF from (c))	✓ 1

Common error: Giving the bearing of P from R as 112.6° − 180° or 22.6°. A back bearing differs by 180°, and the answer must lie between 0° and 360°.

+ Add to test✓ In your test220IGMC3-6-11r40-08Sine rule, cosine rule, trigonometric graphs and 3D problems Paper CMedium6 marks

Calculator

The roof truss of a coffee warehouse is an isosceles triangle. The base is 9.6 m long and each of the two equal sloping sides is 7.4 m long.

Diagram not to scale

(a)

Calculate the height of the truss. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the angle between a sloping side and the base. Give your answer correct to 1 decimal place.

(2)

(c)

Calculate the area of the triangle. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{7.4^2-4.8^2}\) or \(\sqrt{31.72}\)	✓ 1
A1 5.63 m	✓ 1
Part (b)
M1 \(\cos\theta=\frac{4.8}{7.4}\) or \(\tan\theta=\frac{5.63}{4.8}\)	✓ 1
A1 49.6°	✓ 1
Part (c)
M1 \(\frac12\times9.6\times\) their height from (a)	✓ 1
A1 27.0 m²	✓ 1

Common error: Using 7.4 as the height or 9.6 as the half-base. Mark the right angle and the half-base of 4.8 m on your sketch before choosing a formula.

+ Add to test✓ In your test221IGMC3-6-11r40-11Bearings Paper CMedium7 marks

No calculator

Three villages A, B and C are such that AB = BC. The bearing of B from A is 072°. The bearing of C from B is 148°.

AB = BC. Diagram not to scale

(a)

Write down the bearing of A from B.

(1)

(b)

Find angle ABC.

(2)

(c)

Find angle BAC. Give a reason for your method.

(2)

(d)

Find the bearing of C from A.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 252°	✓ 1
Part (b)
M1 252° − 148°	✓ 1
A1 104°	✓ 1
Part (c)
M1 \((180^\circ-104^\circ)\div2\)	✓ 1
A1 38° because triangle ABC is isosceles with AB = BC, so the base angles are equal	✓ 1
Part (d)
M1 72° + 38°	✓ 1
A1 110°	✓ 1

Exam tip: Without a diagram, work out the back bearing first, then use the fact that the angles in a triangle add up to 180°. A quick sketch with a north line at each point prevents sign errors.

+ Add to test✓ In your test222IGMC3-6-11r40-14Bearings Paper CMedium4 marks

Calculator

The gallery of a lighthouse is 35 m above sea level. From the gallery, the angle of depression of a boat on the sea is 28°. Assume the sea is level and the boat is in line with the base of the lighthouse.

Gallery at the top of the lighthouse. Diagram not to scale

(a)

Calculate the horizontal distance from the boat to the base of the lighthouse. Give your answer correct to 3 significant figures.

(2)

(b)

The boat then moves 20 m directly towards the lighthouse. Calculate the new angle of depression. Give your answer correct to 1 decimal place. (You may use the unrounded or the 3 significant figure value from (a); both give the same answer.)

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\tan28^\circ=\frac{35}{d}\) or \(d=\frac{35}{\tan28^\circ}\)	✓ 1
A1 65.8 m	✓ 1
Part (b)
M1 \(\tan\theta=\frac{35}{65.8-20}\)	✓ 1
A1 37.4°	✓ 1

Common error: Using the angle of depression as the angle at the top of the right-angled triangle. The angle of depression equals the angle of elevation from the boat, so it sits at the boat, opposite the 35 m side.

+ Add to test✓ In your test223IGMC3-6-10r45-03Bearings Paper CMedium7 marks

Calculator

A search boat leaves harbour H and sails 12 km on a bearing of 060° to point P. It then sails 9 km on a bearing of 150° to point Q.

Diagram not to scale

(a)

Show that angle HPQ = 90°.

(2)

(b)

Calculate the distance HQ.

(2)

(c)

Calculate the bearing of Q from H. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 the bearing of H from P is 060° + 180° = 240°	✓ 1
B1 angle HPQ = 240° − 150° = 90°	✓ 1
Part (b)
M1 12² + 9² (= 225)	✓ 1
A1 15 km	✓ 1
Part (c)
M1 tan(PHQ) = 9 ÷ 12	✓ 1
A1 angle PHQ = 36.9°	✓ 1
A1 bearing = 060° + 36.9° = 096.9°	✓ 1

Common error: Draw a north line at every point where a bearing is used; a back bearing is the forward bearing plus or minus 180°.

+ Add to test✓ In your test224IGMC3-6-10r45-04Pythagoras' theorem and right-angled trigonometry Paper CMedium7 marks

Calculator

In a data centre, a straight cable tray rises from the floor at an angle of 18° to the horizontal. The tray is 14.5 m long.

Diagram not to scale

(a)

Calculate the vertical height gained by the tray. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the horizontal distance covered by the tray. Give your answer correct to 3 significant figures.

(2)

(c)

A second straight tray rises 3.2 m over a horizontal distance of 11.0 m. Calculate the angle this tray makes with the horizontal. Give your answer correct to 1 decimal place.

(2)

(d)

State which tray is steeper. Give a reason.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 14.5 × sin 18°	✓ 1
A1 4.48 m	✓ 1
Part (b)
M1 14.5 × cos 18°	✓ 1
A1 13.8 m	✓ 1
Part (c)
M1 tan⁻¹(3.2 ÷ 11.0)	✓ 1
A1 16.2°	✓ 1
Part (d)
B1 the first tray, because 18° is greater than 16.2° (ECF from (c))	✓ 1

Exam tip: Check your calculator is in degree mode before using sin, cos or tan with angles in degrees.

+ Add to test✓ In your test225IGMC3-6-10r45-07Pythagoras' theorem and right-angled trigonometry Paper CMedium6 marks

Calculator

An archaeologist stands on level ground 35 m from the base of a vertical stone wall. Her eyes are 1.6 m above the ground. The angle of elevation of the top of the wall from her eyes is 28°.

Diagram not to scale

(a)

Calculate the height of the wall above the ground. Give your answer correct to 3 significant figures.

(3)

(b)

She walks towards the wall until the angle of elevation of the top is 45°. Calculate how far she walks. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 height above eye level = 35 × tan 28°	✓ 1
A1 18.6 m	✓ 1
A1 18.6 + 1.6 = 20.2 m	✓ 1
Part (b)
B1 at 45° the distance from the wall equals the height above her eyes, 18.6 m	✓ 1
M1 35 − 18.6	✓ 1
A1 16.4 m	✓ 1

Common error: The angle of elevation is measured from the observer's eye level, so add the eye height to the height found.

+ Add to test✓ In your test226IGMC3-6-10r45-08Pythagoras' theorem and right-angled trigonometry Paper CMedium8 marks

Calculator

The end wall of a data-centre storage shed is an isosceles triangle. Its base is 9.0 m long and its two equal sloping sides are each 7.5 m long.

Diagram not to scale

(a)

Calculate the height of the end wall.

(3)

(b)

Calculate the angle between a sloping side and the base. Give your answer correct to 1 decimal place.

(2)

(c)

The end wall is covered with panels costing €48 per square metre. Calculate the cost of the panels.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 half of the base = 4.5 m	✓ 1
M1 7.5² − 4.5² (= 36)	✓ 1
A1 6.0 m	✓ 1
Part (b)
M1 cos⁻¹(4.5 ÷ 7.5) (or tan⁻¹(6.0 ÷ 4.5), sin⁻¹(6.0 ÷ 7.5))	✓ 1
A1 53.1°	✓ 1
Part (c)
M1 area = ½ × 9.0 × 6.0	✓ 1
A1 27 m²	✓ 1
A1 €1296	✓ 1

Exam tip: Drawing the perpendicular height in an isosceles triangle splits it into two identical right-angled triangles.

+ Add to test✓ In your test227IGMC3-6-10r45-09Bearings Paper CMedium6 marks

Calculator

A rescue helicopter leaves base B and flies 36 km on a bearing of 125° to point P. It then flies due west until it is due south of B, at point Q.

Diagram not to scale

(a)

Show that the angle between BP and the line from B due south is 55°.

(1)

(b)

Calculate the distance PQ. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the distance BQ. Give your answer correct to 3 significant figures.

(2)

(d)

State the bearing of B from Q.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 180° − 125° = 55°	✓ 1
Part (b)
M1 36 × sin 55°	✓ 1
A1 29.5 km	✓ 1
Part (c)
M1 36 × cos 55°	✓ 1
A1 20.6 km	✓ 1
Part (d)
B1 000°	✓ 1

Common error: A bearing of 125° lies between east (090°) and south (180°), so the angle with the south line is 180° − 125°, not 125° − 90°.

+ Add to test✓ In your test228IGMC3-6-12r45-03Bearings Paper CMedium7 marks

Calculator

A medical helicopter flies from a base A on a bearing of 070° for 30 km to a hospital B. It then flies on a bearing of 160° for 40 km to a second hospital C.

Diagram not to scale

(a)

Show that angle ABC is 90° and calculate the straight-line distance AC.

(3)

(b)

Calculate the bearing of C from A. Give your answer to 1 decimal place.

(3)

(c)

Find the bearing of A from C. Give your answer to 1 decimal place.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 the bearing of A from B is 250°, so angle ABC = 250° − 160° = 90°	✓ 1
M1 AC² = 30² + 40²	✓ 1
A1 AC = 50 km	✓ 1
Part (b)
M1 tan(angle BAC) = 40 ÷ 30	✓ 1
A1 angle BAC = 53.1°	✓ 1
A1 bearing = 070° + 53.1° = 123.1°	✓ 1
Part (c)
B1 303.1° (123.1° + 180°, ECF from (b))	✓ 1

Exam tip: A bearing is measured clockwise from north; for a back bearing add or subtract 180°.

+ Add to test✓ In your test229IGMC3-6-12r45-04Bearings Paper CMedium7 marks

Calculator

An ambulance station S is 8 km due south of a hospital H. A clinic C is 12 km from S on a bearing of 053°.

Diagram not to scale

(a)

Calculate how far C is due east of S. Give your answer to 3 significant figures.

(2)

(b)

Calculate how far C is due north of S. Give your answer to 3 significant figures.

(2)

(c)

Calculate the distance CH. Use the unrounded values from parts (a) and (b) and give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12 × sin 53°	✓ 1
A1 9.58 km	✓ 1
Part (b)
M1 12 × cos 53°	✓ 1
A1 7.22 km	✓ 1
Part (c)
B1 C is 8 − 12 cos 53° = 0.778 km south of H	✓ 1
M1 CH² = (12 sin 53°)² + (8 − 12 cos 53°)²	✓ 1
A1 9.62 km (accept 9.61 if rounded values are used)	✓ 1

Common error: On a bearing, the east distance uses sine and the north distance uses cosine of the bearing angle, because the angle is measured from north.

+ Add to test✓ In your test230IGMC3-6-12r45-05Pythagoras' theorem and right-angled trigonometry Paper CMedium6 marks

Calculator

A helipad marking at a hospital is a triangle with sides of 20 m, 21 m and 29 m.

Diagram not to scale

(a)

Show that the triangle has a right angle.

(2)

(b)

Calculate the angle between the sides of length 20 m and 29 m. Give your answer to 1 decimal place.

(2)

(c)

Calculate the shortest distance from the right-angled corner to the longest side. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 20² + 21² = 841	✓ 1
A1 29² = 841, so the triangle is right-angled (converse of Pythagoras)	✓ 1
Part (b)
M1 cos θ = 20 ÷ 29	✓ 1
A1 46.4°	✓ 1
Part (c)
M1 area = ½ × 20 × 21 = 210 and ½ × 29 × h = 210	✓ 1
A1 14.5 m	✓ 1

Exam tip: To show a triangle is right-angled, show that the sum of the squares of the two shorter sides equals the square of the longest.

+ Add to test✓ In your test231IGMC3-6-12r45-13Pythagoras' theorem and right-angled trigonometry Paper CMedium6 marks

Calculator

The gable end of a hospital shed is an isosceles triangle with two equal sides of 13 m and a horizontal base of 10 m.

Diagram not to scale

(a)

Calculate the vertical height of the gable.

(2)

(b)

Calculate the angle between one sloping side and the base. Give your answer to 1 decimal place.

(2)

(c)

Calculate the area of the gable.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 height² = 13² − 5²	✓ 1
A1 12 m	✓ 1
Part (b)
M1 tan θ = 12 ÷ 5 (or cos θ = 5 ÷ 13)	✓ 1
A1 67.4°	✓ 1
Part (c)
M1 ½ × 10 × 12 (ECF from (a))	✓ 1
A1 60 m²	✓ 1

Exam tip: Split an isosceles triangle down the middle to create two right-angled triangles with half the base.

+ Add to test✓ In your test232IGMC3-6-11r46-05Bearings Paper CMedium5 marks

Calculator

A rescue boat leaves port P and sails 16 km on a bearing of 125° to a buoy A. The point B is due south of P and due west of A.

Diagram not to scale

(a)

Show that angle APB is 55°.

(2)

(b)

Calculate the distance AB. Give your answer to 3 significant figures.

(2)

(c)

The boat returns directly from A to P. Find the bearing of P from A.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing 125° is 125° clockwise from north, so the angle from south is \(180^\circ-125^\circ\)	✓ 1
A1 55° shown	✓ 1
Part (b)
M1 \(16\sin55^\circ\)	✓ 1
A1 13.1 km	✓ 1
Part (c)
B1 \(125+180=305\)°	✓ 1

Common error: To find a return bearing, add or subtract 180° from the outward bearing; do not simply reverse the digits.

+ Add to test✓ In your test233IGMC3-6-11r46-08Bearings Paper CMedium7 marks

Calculator

A search vessel leaves port P and sails 18 km on a bearing of 065° to a point A. It then sails 24 km on a bearing of 155° to a point B.

Diagram not to scale

(a)

Show that angle PAB is 90°.

(2)

(b)

Calculate the distance PB.

(2)

(c)

Calculate the bearing of B from P, giving your answer to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 the bearing of P from A is \(065+180=245\)°	✓ 1
A1 \(245-155=90\)° shown	✓ 1
Part (b)
M1 \(\sqrt{18^2+24^2}\)	✓ 1
A1 30 km	✓ 1
Part (c)
M1 \(\tan(\angle APB)=\frac{24}{18}\), giving \(53.1^\circ\)	✓ 1
M1 adds this angle to the bearing 065°	✓ 1
A1 118.1°	✓ 1

Exam tip: Mark the north line at every turning point and use parallel north lines to find angles in bearing problems.

+ Add to test✓ In your test234IGMC3-6-11r46-10Pythagoras' theorem and right-angled trigonometry Paper CMedium6 marks

Calculator

The cross-section of a fish-farm channel is an isosceles trapezium. The horizontal bottom is 6 m wide, the horizontal water surface is 14 m wide, and the vertical depth is 3.2 m.

Cross-section of the channel. Diagram not to scale

(a)

Show that the horizontal distance covered by each sloping side is 4 m.

(2)

(b)

Calculate the length of one sloping side. Give your answer to 3 significant figures.

(2)

(c)

Calculate the angle between a sloping side and the horizontal. Give your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \((14-6)\div2\)	✓ 1
A1 4 m shown	✓ 1
Part (b)
M1 \(\sqrt{4^2+3.2^2}\)	✓ 1
A1 5.12 m	✓ 1
Part (c)
M1 \(\tan\theta=\frac{3.2}{4}\)	✓ 1
A1 38.7°	✓ 1

Common error: The sloping side is the hypotenuse, so do not use the full 14 m width as the base of the right-angled triangle.

+ Add to test✓ In your test235IGMC3-6-11r46-12Pythagoras' theorem and right-angled trigonometry Paper CMedium6 marks

Calculator

The end wall of a barn is an isosceles triangle. The base is 12 m long and each of the two equal sides is 6.5 m long.

Diagram not to scale

(a)

Calculate the height of the wall above its base.

(2)

(b)

Calculate the angle between a sloping side and the base. Give your answer to 1 decimal place.

(2)

(c)

Calculate the area of the triangular wall.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{6.5^2-6^2}\)	✓ 1
A1 2.5 m	✓ 1
Part (b)
M1 \(\cos\theta=\frac{6}{6.5}\)	✓ 1
A1 22.6°	✓ 1
Part (c)
M1 \(\frac12\times12\times2.5\)	✓ 1
A1 15 m²	✓ 1

Common error: To use Pythagoras in an isosceles triangle, split it into two right-angled triangles using half the base.

+ Add to test✓ In your test236IGMC3-6-03d-12Pythagoras' theorem and right-angled trigonometry Paper CHard11 marks

Calculator

Two vertical towers P and Q stand on level ground, 40 m apart. Tower P is 24 m tall and tower Q is 15 m tall. A straight zip wire runs from the top of P to the top of Q.

Diagram not to scale

(a)

Calculate the length of the zip wire. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate the angle of depression of the zip wire. Give your answer correct to 1 decimal place.

(2)

(c)

A second straight wire runs from the top of P to a point R on the ground, beyond the foot of tower Q and in line with both towers. This wire has an angle of depression of 28°. Calculate the distance from the foot of Q to R. Use unrounded values and give your answer correct to the nearest 0.1 m.

(4)

(d)

Calculate the length of the second wire. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 vertical drop = 24 − 15 = 9 m	✓ 1
M1 √(40² + 9²) = √1681	✓ 1
A1 41.0 m	✓ 1
Part (b)
M1 tan θ = 9 ÷ 40	✓ 1
A1 12.7°	✓ 1
Part (c)
M1 horizontal distance from P to R = 24 ÷ tan 28°	✓ 1
A1 45.1 m (45.1 m)	✓ 1
M1 subtract 40 m	✓ 1
A1 5.1 m	✓ 1
Part (d)
M1 24 ÷ sin 28° (or 24 ÷ tan 28° ÷ cos 28°)	✓ 1
A1 51.1 m	✓ 1

Exam tip: Draw and label a side view of the situation with the right angles marked, then pick out each right-angled triangle separately.

+ Add to test✓ In your test237IGMC3-6-09r6-03Bearings Paper CHard8 marks

Calculator

A cargo plane flies from airport A to airport B, a distance of 180 km on a bearing of 065°. It then flies 100 km due south to airport C. Treat the ground as flat.

Diagram not to scale

(a)

Calculate how far east of A airport B is. Give your answer to 3 significant figures.

(2)

(b)

Calculate how far north of A airport B is. Give your answer to 3 significant figures.

(2)

(c)

Calculate the bearing of C from A. Work with unrounded values and give your answer to 1 decimal place.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses 180 sin 65°	✓ 1
East distance = 163 km	✓ 1
Part (b)
Uses 180 cos 65°	✓ 1
North distance = 76.1 km	✓ 1
Part (c)
C is 23.93 km south of A (100 − 76.07), ECF from (b)	✓ 1
Angle below the east direction: tan θ = 23.93/163.14	✓ 1
θ = 8.3°	✓ 1
Bearing = 90° + 8.3° = 98.3°	✓ 1

Common error: A point south of the start but east of it has a bearing between 090° and 180°; check your final angle is in the correct quadrant.

+ Add to test✓ In your test238IGMC3-6-10r6-07Pythagoras' theorem and right-angled trigonometry Paper CHard7 marks

Calculator

From the top of a vertical cliff 62 m above the sea, an observer sees a boat at an angle of depression of 28°. The boat then sails in a straight line directly towards the base of the cliff until the angle of depression of the boat from the top of the cliff is 50°. Assume the sea is level.

Diagram not to scale

(a)

Explain why the angle of elevation of the top of the cliff from the boat is also 28°.

(1)

(b)

Calculate the original horizontal distance of the boat from the base of the cliff. Give your answer to 3 significant figures.

(2)

(c)

Calculate the horizontal distance of the boat from the base of the cliff when the angle is 50°. Give your answer to 3 significant figures.

(2)

(d)

Calculate the distance the boat sails. Use unrounded values and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Alternate angles between parallel lines (horizontal at the top and the sea surface) are equal	✓ 1
Part (b)
tan 28° = 62/distance	✓ 1
Distance = 117 m	✓ 1
Part (c)
tan 50° = 62/distance	✓ 1
Distance = 52.0 m	✓ 1
Part (d)
Distance sailed = 62/tan 28° − 62/tan 50° (116.61 − 52.02)	✓ 1
Distance = 64.6 m	✓ 1

Common error: Rounding the two distances before subtracting can change the third significant figure; keep full calculator values until the end.

+ Add to test✓ In your test239IGMC3-6-10r7-11Pythagoras' theorem and right-angled trigonometry Paper CHard6 marks

Calculator

Points P and Q are on level ground in a straight line with the base B of a wind turbine tower, with Q between P and B. PQ = 70 m. The angle of elevation of the top of the tower is 28° from P and 41° from Q. Let the height of the tower be h metres.

Diagram not to scale

(a)

Write expressions, in terms of h, for the distances PB and QB.

(2)

(b)

Use your expressions to calculate h. Give your answer correct to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 PB = h ÷ tan 28°	✓ 1
B1 QB = h ÷ tan 41°	✓ 1
Part (b)
M1 PB − QB = 70	✓ 1
M1 h(1 ÷ tan 28° − 1 ÷ tan 41°) = 70	✓ 1
M1 h = 70 ÷ (1.880… − 1.150…)	✓ 1
A1 95.8 m	✓ 1

Exam tip: With two angles of elevation, express the same unknown height in two ways and use the known distance between the viewing points to link them.

+ Add to test✓ In your test240IGMC3-6-09r9-14Pythagoras' theorem and right-angled trigonometry Paper CHard8 marks

Calculator

A bridge tower is 50 m tall and stands vertically on level ground. Two straight cables run from the top of the tower to anchor points A and B on the ground, on opposite sides of the tower and in the same vertical plane. The cable to A makes an angle of 62° with the ground. The cable to B makes an angle of 35° with the ground.

Diagram not to scale

(a)

Calculate the horizontal distance from the base of the tower to A. Give your answer correct to 1 decimal place.

(2)

(b)

Calculate the horizontal distance from the base of the tower to B. Give your answer correct to 1 decimal place.

(2)

(c)

Calculate the distance AB.

(2)

(d)

Calculate the length of the longer cable. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 50 ÷ tan 62°	✓ 1
A1 26.6 m	✓ 1
Part (b)
M1 50 ÷ tan 35°	✓ 1
A1 71.4 m	✓ 1
Part (c)
M1 add the two distances: 26.6 + 71.4 (ECF from (a), (b))	✓ 1
A1 98.0 m	✓ 1
Part (d)
M1 the longer cable is the one at 35°: 50 ÷ sin 35°	✓ 1
A1 87.2 m	✓ 1

Exam tip: A shallower cable angle means a longer cable for the same height, so identify the longer one before calculating.

+ Add to test✓ In your test241IGMC3-6-10r25-06Bearings Paper CHard7 marks

Calculator

A survey boat leaves harbour \(H\) and sails 10 km on a bearing of 060° to a point \(A\). It then sails 14 km on a bearing of 150° to a point \(B\).

Diagram not to scale

(a)

Show that angle \(HAB = 90°\).

(2)

(b)

Calculate the distance \(HB\), to 3 significant figures.

(2)

(c)

Calculate the bearing of \(B\) from \(H\), to the nearest degree.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 the bearing of \(H\) from \(A\) is \(240°\) (or co-interior angles give 60° at \(A\))	✓ 1
B1 \(240° - 150° = 90°\)	✓ 1
Part (b)
M1 \(\sqrt{10^2 + 14^2}\)	✓ 1
A1 17.2 km	✓ 1
Part (c)
M1 \(\tan^{-1}\left(\frac{14}{10}\right) = 54.5°\) for angle \(AHB\)	✓ 1
M1 adds to the bearing 060° (since \(B\) is clockwise of \(HA\))	✓ 1
A1 114°	✓ 1

Exam tip: In multi-leg bearing questions, add or subtract the angle in the triangle from a known bearing, and decide the direction with a sketch.

+ Add to test✓ In your test242IGMC3-6-08r26-12Bearings Paper CHard9 marks

Calculator

A ship carrying coffee sails from port A on a bearing of 120° for 40 km to port B. It then sails due south for 25 km to port C.

Diagram not to scale

(a)

Show that B is 34.6 km east of A, correct to 3 significant figures.

(2)

(b)

Calculate the distance AC. Give your answer correct to 3 significant figures.

(3)

(c)

Calculate the bearing of C from A. Give your answer correct to 1 decimal place.

(3)

(d)

Write down the bearing of A from C.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 40 × sin 60° (or 40 × cos 30°)	✓ 1
A1 34.64..., so 34.6 km, shown	✓ 1
Part (b)
M1 B is 40 × cos 60° = 20 km south of A, so C is 45 km south of A	✓ 1
M1 AC² = 45² + 34.64²	✓ 1
A1 56.8 km	✓ 1
Part (c)
M1 tan(angle between AC and south) = 34.64 ÷ 45	✓ 1
A1 37.6°	✓ 1
A1 180° − 37.6° = 142.4°	✓ 1
Part (d)
B1 322.4°	✓ 1

Exam tip: Split each leg of a journey into east-west and north-south parts, then add the parts for the whole journey.

+ Add to test✓ In your test243IGMC3-6-10r30-08Pythagoras' theorem and right-angled trigonometry Paper CHard4 marks

No calculator

A delivery drone leaves a coffee roastery and flies 10 km on a bearing of 090° and then 24 km on a bearing of 000°. It is then at point D.

(a)

Work out the straight-line distance from the roastery to D.

(2)

(b)

The straight line from the roastery to D makes an angle \(x\) with north. Write down the exact value of \(\tan x\) as a fraction in its simplest form.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{10^2+24^2}=\sqrt{676}\)	✓ 1
A1 26 km	✓ 1
Part (b)
M1 \(\tan x=\frac{10}{24}\) (opposite east, adjacent north)	✓ 1
A1 \(\frac{5}{12}\)	✓ 1

Common error: Draw the two legs as a right-angled triangle with north as one leg before choosing which sides are opposite and adjacent.

+ Add to test✓ In your test244IGMC3-6-11r33-14Pythagoras' theorem and right-angled trigonometry Paper CHard8 marks

Calculator

A ladder of length 6.5 m leans against a vertical wall of a water tank with its foot on level ground 2.5 m from the wall.

(a)

Calculate the height the top of the ladder reaches up the wall.

(2)

(b)

Calculate the angle between the ladder and the ground, correct to 1 decimal place.

(2)

(c)

The ladder slips so that its top moves 0.6 m down the wall, with its foot still on the ground and the ladder still touching the wall. Calculate how far the foot of the ladder moves, correct to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(6.5² − 2.5²)	✓ 1
A1 6 m	✓ 1
Part (b)
M1 cos x = 2.5/6.5	✓ 1
A1 67.4°	✓ 1
Part (c)
M1 new height = 6 − 0.6 = 5.4 m	✓ 1
M1 new distance from wall = √(6.5² − 5.4²)	✓ 1
A1 3.618... m	✓ 1
A1 1.12 m	✓ 1

Exam tip: The ladder length stays the same in both positions, so use it as the hypotenuse each time.

+ Add to test✓ In your test245IGMC3-6-10r40-07Pythagoras' theorem and right-angled trigonometry Paper CHard7 marks

Calculator

A tent has a cross-section in the shape of an isosceles triangle. The base is 2.0 m wide and each sloping side is 1.7 m long.

Diagram not to scale

(a)

Calculate the height of the tent. Give your answer to 3 significant figures.

(3)

(b)

Calculate the angle between a sloping side and the ground. Give your answer to 1 decimal place.

(2)

(c)

Calculate the area of the cross-section. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 splits the triangle into two right-angled triangles with base 1.0 m	✓ 1
M1 \(\sqrt{1.7^2-1.0^2}=\sqrt{1.89}\)	✓ 1
A1 1.37 m	✓ 1
Part (b)
M1 \(\cos x=\dfrac{1.0}{1.7}\)	✓ 1
A1 54.0°	✓ 1
Part (c)
M1 \(\dfrac12\times 2.0\times\) height (accept use of 1.37)	✓ 1
A1 1.37 m²	✓ 1

Common error: The perpendicular height meets the base at its midpoint, so use half the base, not the full 2.0 m, in the right-angled triangle.

+ Add to test✓ In your test246IGMC3-6-12r45-06Pythagoras' theorem and right-angled trigonometry Paper CHard5 marks

Calculator

A paramedic stands on level ground in line with a hospital tower. The angle of elevation of the top of the tower is 28°. She walks 40 m directly towards the tower and the angle of elevation of the top is now 41°. Ignore the paramedic's height.
Let the height of the tower be h metres.

Diagram not to scale

(a)

Write down expressions, in terms of h, for her horizontal distance from the tower at each position.

(2)

(b)

Calculate the height of the tower. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 first position: h ÷ tan 28°	✓ 1
B1 second position: h ÷ tan 41°	✓ 1
Part (b)
M1 h ÷ tan 28° − h ÷ tan 41° = 40	✓ 1
M1 h = 40 ÷ (1 ÷ tan 28° − 1 ÷ tan 41°)	✓ 1
A1 54.8 m	✓ 1

Common error: Do not round the intermediate values for 1 ÷ tan 28° and 1 ÷ tan 41°; keep full calculator values until the final answer.

+ Add to test✓ In your test247IGMC3-6-11r46-13Pythagoras' theorem and right-angled trigonometry Paper CHard6 marks

Calculator

A researcher stands on a cliff-top where her eyes are 42 m above the sea. She watches a seal on the sea. The angle of depression of the seal is 14°. The seal then swims directly away from the cliff, and the angle of depression becomes 8°.

Cliff and sea. Diagram not to scale

(a)

Calculate the horizontal distance from the cliff to the seal when the angle of depression is 14°. Give your answer to the nearest metre.

(2)

(b)

Calculate the horizontal distance from the cliff to the seal when the angle of depression is 8°. Give your answer to the nearest metre.

(2)

(c)

Calculate how far the seal swam. Use the unrounded values from (a) and (b) and give your answer to the nearest metre.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 angle in the triangle at the seal is 14° (alternate angles), so \(\tan14^\circ=\frac{42}{d}\)	✓ 1
A1 168 m	✓ 1
Part (b)
M1 \(\frac{42}{\tan8^\circ}\)	✓ 1
A1 299 m	✓ 1
Part (c)
M1 \(\frac{42}{\tan8^\circ}-\frac{42}{\tan14^\circ}\)	✓ 1
A1 130 m	✓ 1

Common error: The angle of depression is measured down from the horizontal, so it equals the angle of elevation from the seal to the observer.

+ Add to test✓ In your test248IGMC-046-02-01Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EEasy5 marks

Calculator

In triangle ABC, angle BAC = 48°, angle ABC = 63° and BC = 9.4 cm.

Diagram not to scale

(a)

Calculate the length of AC.

(3)

(b)

Calculate the length of AB.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: AC / sin 63° = 9.4 / sin 48°	✓ 1
M1: AC = 9.4 × sin 63° ÷ sin 48°	✓ 1
A1: AC = 11.3 cm (11.27…)	✓ 1
Part (b)
M1: angle ACB = 180° − 48° − 63° = 69°, then AB = 9.4 × sin 69° ÷ sin 48°	✓ 1
A1: AB = 11.8 cm (11.808…)	✓ 1

Exam tip: Label each side with the angle opposite it before using the sine rule. BC is opposite angle A, AC is opposite angle B and AB is opposite angle C.

+ Add to test✓ In your test249IGMC-046-02-02Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EEasy5 marks

Calculator

In triangle PQR, PQ = 7.2 cm, PR = 5.8 cm and angle QPR = 112°.

Diagram not to scale

(a)

Calculate the length of QR.

(3)

(b)

Calculate the area of triangle PQR.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: QR² = 7.2² + 5.8² − 2 × 7.2 × 5.8 × cos 112°	✓ 1
A1: QR² = 116.7(67…)	✓ 1
A1: QR = 10.8 cm	✓ 1
Part (b)
M1: ½ × 7.2 × 5.8 × sin 112°	✓ 1
A1: 19.4 cm² (19.35…)	✓ 1

Common error: cos 112° is negative, so subtracting 2bc cos A makes QR² larger than 7.2² + 5.8². An answer for QR below 9.24 cm means the sign has been lost.

+ Add to test✓ In your test250IGMC-046-02-03Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EEasy4 marks

Calculator

Triangle DEF has sides DE = 8 cm, EF = 11 cm and DF = 13 cm.

Diagram not to scale

(a)

Calculate the size of the largest angle of the triangle. Give your answer correct to 1 decimal place.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: the largest angle is angle DEF, which is opposite the longest side DF	✓ 1
M1: cos E = (8² + 11² − 13²) ÷ (2 × 8 × 11)	✓ 1
M1: cos E = 16/176 (= 0.0909…)	✓ 1
A1: angle DEF = 84.8°	✓ 1

Exam tip: The largest angle is always opposite the longest side. The side that is subtracted in the numerator must be the side opposite the angle you want.

+ Add to test✓ In your test251IGMC-046-02-04Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EEasy4 marks

No calculator

The graph of y = sin x is drawn for 0° ≤ x ≤ 360°.
You are given that sin 30° = 0.5.

(a)

Write down the coordinates of the maximum point of the graph.

(1)

(b)

Write down all the values of x where the graph crosses or touches the x-axis.

(1)

(c)

Solve the equation sin x = −0.5 for 0° ≤ x ≤ 360°.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: (90, 1)	✓ 1
Part (b)
B1: 0°, 180°, 360° (all three)	✓ 1
Part (c)
B1: x = 210°	✓ 1
B1: x = 330°	✓ 1

Exam tip: Use the symmetry of the sine curve. The graph is below the x-axis for 180° < x < 360°, so the solutions are 180° + 30° and 360° − 30°.

+ Add to test✓ In your test252IGMC-046-02-05Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EEasy6 marks

Calculator

Give all answers in this question correct to 1 decimal place.

(a)

Solve tan x = 3 for 0° ≤ x ≤ 360°.

(3)

(b)

Solve 5 cos x + 2 = 0 for 0° ≤ x ≤ 360°.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: x = tan⁻¹ 3	✓ 1
A1: x = 71.6°	✓ 1
B1: x = 251.6° (their 71.6° + 180°)	✓ 1
Part (b)
M1: cos x = −0.4	✓ 1
A1: x = 113.6°	✓ 1
B1: x = 246.4° (360° − their 113.6°)	✓ 1

Common error: Stopping at the calculator value loses the second solution. The graph of tan x repeats every 180°. The graph of cos x is symmetrical about x = 180°.

+ Add to test✓ In your test253IGMC3-6-03d-11Pythagoras' theorem and right-angled trigonometry Paper EEasy6 marks

Calculator

A triangular plot of land has two sides of length 14 m and 11 m. The angle between these two sides is 72°.

Diagram not to scale

(a)

Calculate the area of the plot. Give your answer correct to 3 significant figures.

(2)

(b)

Use the cosine rule to calculate the length of the third side. Give your answer correct to 3 significant figures.

(3)

(c)

Fencing is fitted around the whole boundary of the plot. Calculate the total length of fencing needed. Use the unrounded third side and give your answer correct to 3 significant figures.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 area = ½ × 14 × 11 × sin 72°	✓ 1
A1 73.2 m²	✓ 1
Part (b)
M1 c² = 14² + 11² − 2 × 14 × 11 × cos 72°	✓ 1
M1 c² = 221.8	✓ 1
A1 14.9 m	✓ 1
Part (c)
B1 39.9 m (25 + third side, ECF from (b))	✓ 1

Common error: In the cosine rule the angle must be the one between the two known sides. Put the whole product 2ab cos C into your calculator in one go.

+ Add to test✓ In your test254IGMC3-6-09r23-09Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EEasy5 marks

No calculator

The sound wave from a speaker is modelled by y = 3 sin 2x, where x is in degrees and 0° ≤ x ≤ 180°.

(a)

Write down the amplitude of the wave.

(1)

(b)

Write down the period of the wave.

(1)

(c)

Solve 3 sin 2x = 1.5 for 0° ≤ x ≤ 180°.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 3	✓ 1
Part (b)
B1 180°	✓ 1
Part (c)
M1 sin 2x = 0.5	✓ 1
A1 2x = 30° or 150°	✓ 1
A1 x = 15° and x = 75°	✓ 1

Exam tip: For y = a sin bx the period is 360° divided by b, so solve for 2x before dividing by 2.

+ Add to test✓ In your test255IGMC3-6-11r40-07Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EEasy5 marks

Calculator

Two ground stations, A and B, are 440 km apart on level ground. A probe in low orbit is at point C, in the same vertical plane as the two stations. In triangle ABC, angle CAB = 52° and angle CBA = 71°.

Diagram not to scale

(a)

Find angle ACB.

(1)

(b)

Calculate the distance AC. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the distance BC. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 57°	✓ 1
Part (b)
M1 \(\frac{AC}{\sin71^\circ}=\frac{440}{\sin57^\circ}\)	✓ 1
A1 496 km	✓ 1
Part (c)
M1 \(\frac{BC}{\sin52^\circ}=\frac{440}{\sin57^\circ}\)	✓ 1
A1 413 km	✓ 1

Exam tip: In the sine rule each side is paired with the angle opposite it. Find the missing angle first so that the side you know has its opposite angle.

+ Add to test✓ In your test256IGMC-046-01-07Bearings Paper EMedium8 marks

Calculator

A ship sails 40 km from port P on a bearing of 060° to a point Q. It then sails 25 km on a bearing of 170° to a point R.

Diagram not to scale

(a)

Show that angle PQR = 70°.

(2)

(b)

Calculate the distance PR.

(3)

(c)

Calculate the bearing of R from P.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: bearing of P from Q = 240°	✓ 1
B1: 240 − 170 = 70°	✓ 1
Part (b)
M1: PR² = 40² + 25² − 2 × 40 × 25 × cos 70°	✓ 1
M1: PR² = 1540.9…	✓ 1
A1: 39.3 km	✓ 1
Part (c)
M1: sin QPR = 25 sin 70° ÷ their PR	✓ 1
A1: angle QPR = 36.8°	✓ 1
A1: bearing = 60 + 36.8 = 096.8°	✓ 1

Exam tip: Keep the unrounded value of PR in your calculator memory for part (c). Using 39.3 instead can change the final bearing in the first decimal place.

+ Add to test✓ In your test257IGMC-046-01-09Bearings Paper EMedium6 marks

Calculator

Point B is 8 km from point A on a bearing of 110°. A radio tower T is on a bearing of 035° from A and on a bearing of 340° from B.

Diagram not to scale

(a)

Find angle TAB.

(1)

(b)

Find angle ABT.

(2)

(c)

Calculate the distance AT.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: 110 − 35 = 75°	✓ 1
Part (b)
B1: bearing of A from B = 290°	✓ 1
B1: angle ABT = 340 − 290 = 50°	✓ 1
Part (c)
B1: angle ATB = 55°	✓ 1
M1: AT = 8 sin 50° ÷ sin 55°	✓ 1
A1: 7.48 km	✓ 1

Common error: In the sine rule, each side must be paired with the angle opposite it. AT is opposite angle B (50°), and AB is opposite angle T (55°).

+ Add to test✓ In your test258IGMC-046-01-11Bearings Paper EMedium7 marks

Calculator

A light aircraft flies 320 km from airport A on a bearing of 300° to B. It then flies 210 km on a bearing of 045° to C. It then flies directly back from C to A at an average speed of 240 km/h.

Diagram not to scale

(a)

Find angle ABC.

(2)

(b)

Calculate the distance AC.

(3)

(c)

Calculate the time taken to fly from C to A. Give your answer in minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: bearing of A from B = 300 − 180 = 120°	✓ 1
B1: angle ABC = 120 − 45 = 75°	✓ 1
Part (b)
M1: AC² = 320² + 210² − 2 × 320 × 210 × cos 75°	✓ 1
M1: AC = √(111 714.7…) or √(146 500 − 34 785.3…)	✓ 1
A1: AC = 334 km (334.2…)	✓ 1
Part (c)
M1: their AC ÷ 240 × 60	✓ 1
A1: 83.5 to 83.6 minutes (follow through from their AC)	✓ 1

Common error: 1.39 hours is not 1 hour 39 minutes. To convert hours to minutes, multiply the whole decimal number of hours by 60.

+ Add to test✓ In your test259IGMC-046-00-10Pythagoras' theorem and right-angled trigonometry Paper EMedium6 marks

No calculator

Triangle LMN has a right angle at M.
LN = 10 cm and angle MLN = 30°.
You may use sin 30° = \(\frac{1}{2}\) and cos 30° = \(\frac{\sqrt{3}}{2}\).

Diagram not to scale

(a)

Find MN.

(2)

(b)

Find LM, giving your answer in surd form.

(2)

(c)

Find the exact area of triangle LMN.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 10 × sin 30°	✓ 1
A1: 5 cm	✓ 1
Part (b)
M1: 10 × cos 30°	✓ 1
A1: 5√3 cm	✓ 1
Part (c)
M1: ½ × 5 × 5√3, or ½ × their MN × their LM	✓ 1
A1: \(\frac{25\sqrt{3}}{2}\) cm², or an exact equivalent such as 12.5√3 cm²	✓ 1

Common error: Giving 8.66 cm or 21.7 cm² does not answer the question. When an exact answer is asked for, leave √3 in the answer.

+ Add to test✓ In your test260IGMC-046-02-06Bearings Paper EMedium8 marks

Calculator

A boat leaves harbour H and sails 14 km on a bearing of 065° to a buoy A. It then sails 9 km on a bearing of 150° to a lighthouse B.

Diagram not to scale

(a)

Show that angle HAB = 95°.

(2)

(b)

Calculate the distance HB.

(3)

(c)

Calculate the bearing of B from H.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: the bearing of H from A is 065° + 180° = 245°	✓ 1
B1: angle HAB = 245° − 150° = 95°	✓ 1
Part (b)
M1: HB² = 14² + 9² − 2 × 14 × 9 × cos 95°	✓ 1
A1: HB² = 298.9(6…)	✓ 1
A1: HB = 17.3 km (17.29…)	✓ 1
Part (c)
M1: sin AHB / 9 = sin 95° / their 17.29	✓ 1
A1: angle AHB = 31.2° (31.23…)	✓ 1
A1: bearing = 065° + 31.2° = 096.2°	✓ 1

Exam tip: Draw a North line at every point. Using a back bearing (add or subtract 180°) is usually the quickest way to find the angle inside the triangle.

+ Add to test✓ In your test261IGMC-046-02-07Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium7 marks

Calculator

ABCDEFGH is a cuboid. The base ABCD is a horizontal rectangle with AB = 12 cm and BC = 5 cm. The vertical edges are AE, BF, CG and DH, and each is 7 cm long.

Diagram not to scale

(a)

Calculate the length of the diagonal AC.

(2)

(b)

Calculate the length of the diagonal AG.

(2)

(c)

Calculate the angle between AG and the base ABCD.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: AC² = 12² + 5²	✓ 1
A1: AC = 13 cm	✓ 1
Part (b)
M1: AG² = 13² + 7² (= 218)	✓ 1
A1: AG = 14.8 cm (14.76…)	✓ 1
Part (c)
B1: the required angle is angle GAC	✓ 1
M1: tan GAC = 7/13 (or an equivalent ratio using AG)	✓ 1
A1: 28.3°	✓ 1

Exam tip: The angle between a line and a plane is the angle between the line and its projection onto the plane. Here the projection of AG onto the base is AC, because G is directly above C.

+ Add to test✓ In your test262IGMC-046-02-08Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium7 marks

Calculator

In triangle XYZ, XY = 10.5 cm, YZ = 8.2 cm and angle YXZ = 41°. Angle XZY is obtuse.

Diagram not to scale

(a)

Calculate the size of angle XZY.

(4)

(b)

Calculate the area of triangle XYZ.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: sin Z / 10.5 = sin 41° / 8.2	✓ 1
M1: sin Z = 10.5 × sin 41° ÷ 8.2 (= 0.840…)	✓ 1
A1: acute value 57.1° (57.14…)	✓ 1
B1: angle XZY = 180° − 57.1° = 122.9°	✓ 1
Part (b)
M1: angle XYZ = 180° − 41° − 122.9° = 16.1° (16.14…)	✓ 1
M1: ½ × 10.5 × 8.2 × sin(their angle Y)	✓ 1
A1: 12.0 cm² (11.97…)	✓ 1

Common error: Your calculator only gives the acute angle from sin⁻¹. If the angle is obtuse you must subtract from 180°. Keep full accuracy for angle Y, because rounding it to 16.1° leads to 11.9 cm².

+ Add to test✓ In your test263IGMC-046-02-09Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium7 marks

No calculator

In triangle KLM, KL = 6 cm, KM = 10 cm and angle LKM = 60°.
You are given that cos 60° = ½ and sin 60° = \(\frac{\sqrt{3}}{2}\).

Diagram not to scale

(a)

Find the exact length of LM, giving your answer in the form \(a\sqrt{b}\), where a and b are integers.

(3)

(b)

Find the exact area of triangle KLM.

(2)

(c)

Find the exact perpendicular distance from K to the line LM. Give your answer with a rational denominator.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: LM² = 6² + 10² − 2 × 6 × 10 × ½	✓ 1
A1: LM² = 76	✓ 1
A1: LM = 2√19 cm	✓ 1
Part (b)
M1: ½ × 6 × 10 × \(\frac{\sqrt{3}}{2}\)	✓ 1
A1: 15√3 cm²	✓ 1
Part (c)
M1: ½ × 2√19 × h = 15√3	✓ 1
A1: h = \(\frac{15\sqrt{57}}{19}\) cm	✓ 1

Exam tip: The area of a triangle is the same whichever side you use as the base. This lets you find a perpendicular height from an area you already know.

+ Add to test✓ In your test264IGMC-046-02-10Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium6 marks

Calculator

The curve y = a sin x + b, where a > 0, is drawn for 0° ≤ x ≤ 360°. The maximum value of y is 7 and the minimum value of y is −1.

(a)

Find the value of a and the value of b.

(2)

(b)

Find the value of y when x = 210°.

(1)

(c)

Find the values of x where the curve crosses the x-axis. Give your answers correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: a = 4	✓ 1
B1: b = 3	✓ 1
Part (b)
B1: y = 4 × (−0.5) + 3 = 1	✓ 1
Part (c)
M1: 4 sin x + 3 = 0, so sin x = −0.75	✓ 1
A1: x = 228.6°	✓ 1
A1: x = 311.4°	✓ 1

Exam tip: The amplitude is half the distance between the maximum and the minimum. The central line, y = b, lies halfway between them.

+ Add to test✓ In your test265IGMC-046-02-11Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium7 marks

Calculator

VPQRS is a pyramid with a horizontal square base PQRS of side 8 cm. The vertex V is vertically above M, the centre of the square. Each sloping edge (VP, VQ, VR and VS) is 11 cm long.

Diagram not to scale

(a)

Calculate the height VM of the pyramid.

(3)

(b)

Calculate the angle between the edge VP and the base PQRS.

(2)

(c)

Calculate the volume of the pyramid. [Volume of a pyramid = ⅓ × base area × height]

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: PM = ½ × √(8² + 8²) = 4√2 (= 5.657…)	✓ 1
M1: VM² = 11² − (4√2)² (= 89)	✓ 1
A1: VM = 9.43 cm	✓ 1
Part (b)
M1: cos VPM = 5.657 / 11 (or an equivalent ratio)	✓ 1
A1: 59.1°	✓ 1
Part (c)
M1: ⅓ × 64 × their 9.43	✓ 1
A1: 201 cm³ (201.2…)	✓ 1

Common error: Using half the side (4 cm) instead of half the diagonal for PM gives the wrong right-angled triangle. The edge VP meets the base at a corner, not at the midpoint of a side.

+ Add to test✓ In your test266IGMC3-6-03d-08Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium6 marks

Calculator

Points A and B are on one straight bank of a river, 85 m apart. A tree T stands on the opposite bank. Angle TAB = 62° and angle TBA = 71°.

Diagram not to scale

(a)

Find angle ATB.

(1)

(b)

Use the sine rule to calculate the distance AT. Give your answer correct to 3 significant figures.

(3)

(c)

Calculate the shortest distance from T to the line of the bank. Use the unrounded value of AT and give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 47°	✓ 1
Part (b)
M1 AT ÷ sin 71° = 85 ÷ sin 47° (ECF from (a))	✓ 1
M1 AT = 85 × sin 71° ÷ sin 47°	✓ 1
A1 109.9 m	✓ 1
Part (c)
M1 width = AT × sin 62° (ECF from (b))	✓ 1
A1 97.0 m	✓ 1

Exam tip: In the sine rule, each side is paired with the angle directly opposite it. Check that the 85 m side is paired with the 47° angle.

+ Add to test✓ In your test267IGMC3-6-03d-13Pythagoras' theorem and right-angled trigonometry Paper EMedium6 marks

Calculator

A rectangular crate used to ship a telescope has a horizontal base measuring 2.4 m by 1.8 m, and a height of 1.5 m. A straight rod is placed inside, running from one bottom corner to the opposite top corner.

Diagram not to scale

(a)

Calculate the length of the diagonal of the base of the crate.

(2)

(b)

Calculate the length of the rod. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the angle between the rod and the base of the crate. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(2.4² + 1.8²) = √9	✓ 1
A1 3.0 m	✓ 1
Part (b)
M1 √(3.0² + 1.5²) (ECF from (a))	✓ 1
A1 3.35 m	✓ 1
Part (c)
M1 tan θ = 1.5 ÷ 3.0 (or sin θ = 1.5 ÷ 3.35, ECF from (a) or (b))	✓ 1
A1 26.6°	✓ 1

Common error: The angle between a line and a plane is measured to the line's projection on the plane, here the base diagonal, not to an edge of the base.

+ Add to test✓ In your test268IGMC3-6-10r5-09Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium7 marks

Calculator

Three wells A, B and C supply a village. AB = 420 m, angle CAB = 48° and angle ABC = 71°.

Diagram not to scale

(a)

Find angle ACB.

(1)

(b)

Calculate BC, to 3 significant figures.

(3)

(c)

Calculate AC, to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 61°	✓ 1
Part (b)
M1 BC/sin 48° = 420/sin 61°	✓ 1
M1 BC = 420 × sin 48° ÷ sin 61°	✓ 1
A1 357 m	✓ 1
Part (c)
M1 AC/sin 71° = 420/sin 61°	✓ 1
M1 AC = 420 × sin 71° ÷ sin 61°	✓ 1
A1 454 m	✓ 1

Exam tip: Find the third angle first so the sine rule can pair every side with its opposite angle.

+ Add to test✓ In your test269IGMC3-6-10r5-10Pythagoras' theorem and right-angled trigonometry Paper EMedium5 marks

Calculator

In an online game, two players are at points A and B. A beacon O is 85 m from A and 120 m from B, and angle AOB = 62°.

Diagram not to scale

(a)

Calculate the distance AB, to 3 significant figures.

(3)

(b)

Calculate the area of triangle AOB, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 AB² = 85² + 120² − 2 × 85 × 120 × cos 62°	✓ 1
M1 evaluate (AB² = 12048…)	✓ 1
A1 110 m	✓ 1
Part (b)
M1 ½ × 85 × 120 × sin 62°	✓ 1
A1 4500 m²	✓ 1

Common error: The angle in the cosine rule must be the one between the two known sides.

+ Add to test✓ In your test270IGMC3-6-10r5-11Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium8 marks

Calculator

The concentration, C mg/L, of a drug in the blood t hours after a dose is modelled by C = 12 + 5 sin(30t)°, for 0 ≤ t ≤ 12.

(a)

Write down the maximum and minimum concentrations.

(2)

(b)

Write down the period of the model.

(1)

(c)

Find the times when the concentration is 9.5 mg/L.

(3)

(d)

Calculate the concentration when t = 8, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 maximum 17 mg/L	✓ 1
B1 minimum 7 mg/L	✓ 1
Part (b)
B1 12 hours	✓ 1
Part (c)
M1 sin(30t) = −0.5	✓ 1
M1 30t = 210 and 330	✓ 1
A1 t = 7 and t = 11	✓ 1
Part (d)
M1 12 + 5 sin 240°	✓ 1
A1 7.67 mg/L	✓ 1

Common error: Make sure the calculator is in degree mode, since the argument 30t is in degrees.

+ Add to test✓ In your test271IGMC3-6-09r6-08Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium7 marks

Calculator

A triangular field ABC has side BC = 240 m, angle BAC = 51° and angle ABC = 62°. Side BC runs along a river and needs no fence.

Diagram not to scale

(a)

Find angle ACB.

(1)

(b)

Calculate the length of AC. Give your answer to 3 significant figures.

(3)

(c)

Fencing costs €4.80 per metre. Calculate the cost of fencing sides AB and AC. Use unrounded lengths and give your answer to the nearest euro.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
180° − 51° − 62° = 67°	✓ 1
Part (b)
Sine rule: AC/sin 62° = 240/sin 51°	✓ 1
Rearranges to AC = 240 sin 62°/sin 51°	✓ 1
AC = 273 m	✓ 1
Part (c)
AB = 240 sin 67°/sin 51° = 284.3 m	✓ 1
Total length = 556.9 m, cost = total × 4.80	✓ 1
Cost = €2673	✓ 1

Exam tip: In the sine rule each side must be paired with the sine of the angle opposite it.

+ Add to test✓ In your test272IGMC3-6-09r6-09Pythagoras' theorem and right-angled trigonometry Paper EMedium6 marks

Calculator

A grain store is a cuboid. Its rectangular floor ABCD has AB = 12 m and BC = 5 m, and the walls are 4 m high. E, F, G and H are the points directly above A, B, C and D respectively. A cable is stretched in a straight line from A to G.

Diagram not to scale

(a)

Show that the diagonal AC of the floor is 13 m.

(2)

(b)

Calculate the length of the cable AG. Give your answer to 3 significant figures.

(2)

(c)

Calculate the angle between AG and the floor. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
AC² = 12² + 5² = 169	✓ 1
AC = 13 m	✓ 1
Part (b)
AG² = 13² + 4² = 185 (right angle at C)	✓ 1
AG = 13.6 m	✓ 1
Part (c)
The angle is GAC; tan GAC = 4/13	✓ 1
Angle = 17.1°	✓ 1

Common error: The angle between a line and a plane is measured to the line's projection on the plane (here AC), not to a floor edge such as AB.

+ Add to test✓ In your test273IGMC3-6-09r6-10Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium5 marks

Calculator

The displacement y cm of a point on a vibrating string is modelled by y = 1.8 sin 2x°, where x is measured in degrees.

(a)

Write down the amplitude of the graph.

(1)

(b)

Write down the period of the graph in degrees.

(1)

(c)

Find the largest value of x, with 0 ≤ x ≤ 360, for which y = 0.9.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Amplitude = 1.8 cm	✓ 1
Part (b)
Period = 360°/2 = 180°	✓ 1
Part (c)
sin 2x = 0.5 so 2x = 30° or 150° (and add 360°)	✓ 1
The largest 2x in the range up to 720° is 360° + 150° = 510°	✓ 1
x = 255	✓ 1

Exam tip: When the angle inside the sine is a multiple of x, extend the range of the inside angle (here 0 to 720) before solving.

+ Add to test✓ In your test274IGMC3-6-10r6-10Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium6 marks

Calculator

Two monitoring stations A and B are 5.2 km apart on level ground. A vent V on the same level ground is seen from the stations so that angle VAB = 58° and angle VBA = 71°.

Diagram not to scale

(a)

Find angle AVB.

(1)

(b)

Calculate the distance AV. Give your answer to 3 significant figures.

(3)

(c)

Calculate the distance BV. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
180° − 58° − 71° = 51°	✓ 1
Part (b)
Sine rule: AV/sin 71° = 5.2/sin 51°	✓ 1
Rearranges to AV = 5.2 sin 71°/sin 51°	✓ 1
AV = 6.33 km	✓ 1
Part (c)
Sine rule: BV/sin 58° = 5.2/sin 51°	✓ 1
BV = 5.67 km	✓ 1

Exam tip: The side opposite the largest angle is the longest side, which gives a quick check on your two distances.

+ Add to test✓ In your test275IGMC3-6-10r6-12Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium7 marks

Calculator

A theatre set piece is a solid pyramid. Its base ABCD is a horizontal rectangle with AB = 12 m and BC = 9 m. The vertex V is 10 m vertically above M, the centre of the rectangle.

Diagram not to scale

(a)

Show that the diagonal AC is 15 m.

(2)

(b)

Write down the length AM.

(1)

(c)

Calculate the length of the sloping edge VA.

(2)

(d)

Calculate the angle between VA and the base ABCD. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
AC² = 12² + 9² = 225	✓ 1
AC = 15 m	✓ 1
Part (b)
AM = 7.5 m	✓ 1
Part (c)
VA² = 7.5² + 10² = 156.25	✓ 1
VA = 12.5 m	✓ 1
Part (d)
The angle is VAM; tan VAM = 10/7.5	✓ 1
Angle = 53.1°	✓ 1

Common error: The foot of the vertical height is the centre of the base, so use half the diagonal, not the whole diagonal.

+ Add to test✓ In your test276IGMC3-6-10r6-14Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium5 marks

Calculator

A lava flow covers a triangular region PQR of level ground. PQ = 3.8 km, PR = 5.1 km and angle QPR = 52°.

Diagram not to scale

(a)

Calculate the area of the region. Give your answer to 3 significant figures.

(2)

(b)

Calculate the length of QR. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses area = ½ × 3.8 × 5.1 × sin 52°	✓ 1
Area = 7.64 km²	✓ 1
Part (b)
Cosine rule: QR² = 3.8² + 5.1² − 2 × 3.8 × 5.1 × cos 52°	✓ 1
QR² = 16.59	✓ 1
QR = 4.07 km	✓ 1

Exam tip: The angle in ½ab sin C must be the angle between the two sides used.

+ Add to test✓ In your test277IGMC3-6-09r7-05Pythagoras' theorem and right-angled trigonometry Paper EMedium5 marks

Calculator

Three marker posts P, Q and R are fixed on level ground at an archaeological site. PQ = 38.0 m, PR = 52.0 m and angle QPR = 70°.

Diagram not to scale

(a)

Calculate the distance QR. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate the area of triangle PQR. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cosine rule, QR² = 38² + 52² − 2 × 38 × 52 × cos 70°	✓ 1
M1 correct substitution evaluated, QR² = 2796 (accept 2796 to 2797)	✓ 1
A1 52.9 m	✓ 1
Part (b)
M1 area = ½ × 38 × 52 × sin 70°	✓ 1
A1 928 m²	✓ 1

Exam tip: Use the cosine rule when two sides and the angle between them are known. Check that your answer is plausible: the side opposite the largest angle must be the longest.

+ Add to test✓ In your test278IGMC3-6-09r7-06Pythagoras' theorem and right-angled trigonometry Paper EMedium7 marks

Calculator

The temperature, T °C, in a data hall is modelled by T = 21 + 3 sin(30t)°, where t is the number of hours after midnight and 0 ≤ t ≤ 24.

(a)

Write down the maximum and the minimum temperature predicted by the model.

(2)

(b)

Find the period of the model, in hours.

(2)

(c)

Find the first time after midnight when the temperature is 23 °C. Give your answer in hours correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 maximum 24 °C	✓ 1
B1 minimum 18 °C	✓ 1
Part (b)
M1 360 ÷ 30	✓ 1
A1 12 hours	✓ 1
Part (c)
M1 3 sin(30t) = 2, so sin(30t) = 2/3	✓ 1
M1 30t = sin⁻¹(2/3) = 41.8…°	✓ 1
A1 t = 1.39 hours	✓ 1

Common error: Keep the calculator in degree mode for this model. In radian mode sin⁻¹(2/3) gives 0.730, which is not an angle in degrees.

+ Add to test✓ In your test279IGMC3-6-09r7-13Pythagoras' theorem and right-angled trigonometry Paper EMedium7 marks

Calculator

As a cyclist pedals, the height, h metres, of the right pedal above the ground is modelled by h = 0.40 + 0.18 cos x°, for 0 ≤ x ≤ 360, where x is the angle the crank has turned through from its highest position.

(a)

Find the greatest and the least height of the pedal predicted by the model.

(2)

(b)

State the period of the model, in degrees.

(1)

(c)

Find the values of x for which the pedal is 0.49 m above the ground.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 greatest height 0.58 m	✓ 1
B1 least height 0.22 m	✓ 1
Part (b)
B1 360°	✓ 1
Part (c)
M1 0.18 cos x = 0.09, so cos x = 0.5	✓ 1
A1 x = 60°	✓ 1
M1 second solution from symmetry, 360° − 60°	✓ 1
A1 x = 300°	✓ 1

Common error: A cosine equation in the range 0 to 360 has two solutions when the value is between −1 and 1. Finding only the calculator value loses the second solution.

+ Add to test✓ In your test280IGMC3-6-10r7-01Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium6 marks

Calculator

Three wind turbines A, B and C stand on level ground. AB = 640 m, angle CAB = 58° and angle ABC = 71°.

Diagram not to scale

(a)

Find angle ACB.

(1)

(b)

Calculate the distance BC. Give your answer correct to 3 significant figures.

(3)

(c)

Calculate the distance AC. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 180° − 58° − 71° = 51°	✓ 1
Part (b)
M1 sine rule, BC ÷ sin 58° = 640 ÷ sin 51°	✓ 1
M1 BC = 640 × sin 58° ÷ sin 51°	✓ 1
A1 698 m	✓ 1
Part (c)
M1 AC = 640 × sin 71° ÷ sin 51°	✓ 1
A1 779 m	✓ 1

Exam tip: The sine rule pairs each side with the angle opposite it. Find the third angle first so that you can use the side that is known.

+ Add to test✓ In your test281IGMC3-6-10r7-05Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium6 marks

Calculator

Three turbines A, B and C on a wind farm form a triangle on level ground with AB = 820 m, BC = 950 m and AC = 1130 m.

Diagram not to scale

(a)

Calculate angle ABC. Give your answer correct to 1 decimal place.

(4)

(b)

Calculate the area of triangle ABC. Use the unrounded angle and give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cosine rule, 1130² = 820² + 950² − 2 × 820 × 950 × cos B	✓ 1
M1 cos B = (820² + 950² − 1130²) ÷ (2 × 820 × 950)	✓ 1
M1 cos B = 0.191…	✓ 1
A1 79.0°	✓ 1
Part (b)
M1 area = ½ × 820 × 950 × sin B	✓ 1
A1 382000 m²	✓ 1

Exam tip: Use the cosine rule for an angle when all three sides are known. The angle opposite the longest side is the one you can check against a sketch.

+ Add to test✓ In your test282IGMC3-6-10r7-13Bearings Paper EMedium8 marks

Calculator

A wave machine in a swimming pool makes the height, h metres, of the water surface above its still level vary according to h = 0.15 sin(90t)°, where t is the time in seconds and 0 ≤ t ≤ 4.

(a)

State the amplitude and the period of the wave.

(2)

(b)

Calculate the height of the water surface when t = 1.5. Give your answer correct to 3 significant figures.

(2)

(c)

Find the times when the water surface is 0.075 m above its still level. Give each answer correct to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 amplitude 0.15 m	✓ 1
B1 period 360 ÷ 90 = 4 s	✓ 1
Part (b)
M1 h = 0.15 sin 135°	✓ 1
A1 0.106 m	✓ 1
Part (c)
M1 sin(90t) = 0.5	✓ 1
A1 90t = 30°, so t = 0.333 s	✓ 1
M1 second solution 90t = 180° − 30° = 150°	✓ 1
A1 t = 1.67 s	✓ 1

Common error: For sin x = 0.5 in a given range there are two solutions, 30° and 150°. Check that each solution for 90t gives a value of t that lies inside the time interval.

+ Add to test✓ In your test283IGMC3-6-10r8-09Pythagoras' theorem and right-angled trigonometry Paper EMedium6 marks

Calculator

Two telecoms towers A and B are 7.5 km apart on level ground. A relay C is positioned so that angle CAB = 58° and angle ABC = 71°.

Diagram not to scale

(a)

Find angle ACB.

(1)

(b)

Calculate the distance BC. Give your answer to 3 significant figures.

(3)

(c)

Calculate the shortest distance from C to the line AB, using unrounded values. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 51°	✓ 1
Part (b)
M1 BC/sin 58° = 7.5/sin 51°	✓ 1
M1 BC = 7.5 × sin 58° ÷ sin 51°	✓ 1
A1 8.18 km	✓ 1
Part (c)
M1 BC × sin 71° (or AC × sin 58°)	✓ 1
A1 7.74 km	✓ 1

Exam tip: The sine rule needs a side and its opposite angle; find the third angle first.

+ Add to test✓ In your test284IGMC3-6-10r8-10Bearings Paper EMedium6 marks

Calculator

Two climbing parties leave the same base camp. Party 1 walks 6.4 km on a bearing of 035°. Party 2 walks 5.0 km on a bearing of 112°.

Diagram not to scale

(a)

Write down the angle between the two paths.

(1)

(b)

Calculate the distance between the two parties. Give your answer to 3 significant figures.

(3)

(c)

Calculate the area of the triangle formed by the base camp and the two parties. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 112° − 35° = 77°	✓ 1
Part (b)
M1 cosine rule: 6.4² + 5.0² − 2 × 6.4 × 5.0 × cos 77°	✓ 1
A1 distance² = 51.6	✓ 1
A1 7.18 km	✓ 1
Part (c)
M1 ½ × 6.4 × 5.0 × sin 77°	✓ 1
A1 15.6 km²	✓ 1

Common error: Calculating the cosine rule with the calculator in radian mode; check the mode before using trigonometric functions.

+ Add to test✓ In your test285IGMC3-6-10r8-11Bearings Paper EMedium8 marks

Calculator

The temperature, T °C, of a food-testing oven at time t minutes is modelled by T = 180 + 12 sin(30t°), for 0 ≤ t ≤ 12.

(a)

Write down the maximum and minimum temperatures.

(2)

(b)

Write down the period of the model, in minutes.

(1)

(c)

Calculate the temperature when t = 4. Give your answer to 1 decimal place.

(2)

(d)

Find the first time after t = 0 when the temperature is 186 °C.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 maximum 192 °C	✓ 1
B1 minimum 168 °C	✓ 1
Part (b)
B1 360 ÷ 30 = 12 minutes	✓ 1
Part (c)
M1 180 + 12 sin 120°	✓ 1
A1 190.4 °C	✓ 1
Part (d)
M1 12 sin(30t°) = 6	✓ 1
M1 sin(30t°) = 0.5, so 30t = 30	✓ 1
A1 t = 1 minute	✓ 1

Exam tip: For graph-based models, the number in front of the sine is the amplitude and 360 divided by the coefficient of t gives the period.

+ Add to test✓ In your test286IGMC3-6-09r9-08Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium7 marks

Calculator

Surveyors want the width of a straight river. They mark points A and B on their bank, 85 m apart, and sight a tower at C on the far bank. Angle CAB = 48° and angle ABC = 71°.

Diagram not to scale

(a)

Find angle ACB.

(1)

(b)

Calculate the distance AC. Give your answer correct to 3 significant figures.

(3)

(c)

Use AC = 91.9 m to calculate the perpendicular distance from C to the line AB. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 61°	✓ 1
Part (b)
M1 sine rule: AC ÷ sin 71° = 85 ÷ sin 61°	✓ 1
M1 AC = 85 sin 71° ÷ sin 61°	✓ 1
A1 91.9 m	✓ 1
Part (c)
M1 perpendicular distance = AC × sin 48° (ECF from (b))	✓ 1
M1 91.9 × sin 48°	✓ 1
A1 68.3 m	✓ 1

Exam tip: In the sine rule, each side is paired with the sine of the angle opposite it.

+ Add to test✓ In your test287IGMC3-6-09r9-09Bearings Paper EMedium6 marks

Calculator

Two tracking stations, A and B, are 300 km apart. A probe P is 420 km from A and 560 km from B. Treat A, B and P as the corners of a flat triangle.

Diagram not to scale

(a)

Calculate angle PAB. Give your answer correct to 1 decimal place.

(4)

(b)

Use angle PAB = 100.8° to calculate the area of triangle ABP. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cosine rule: 560² = 300² + 420² − 2 × 300 × 420 × cos A	✓ 1
M1 cos A = (300² + 420² − 560²) ÷ (2 × 300 × 420)	✓ 1
A1 cos A = −0.1873 (−47200 ÷ 252000)	✓ 1
A1 100.8°	✓ 1
Part (b)
M1 area = ½ × 300 × 420 × sin 100.8°	✓ 1
A1 61900 km²	✓ 1

Common error: Forgetting that a negative cosine value gives an obtuse angle; the calculator inverse handles this if the sign is kept.

+ Add to test✓ In your test288IGMC3-6-09r9-10Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium7 marks

Calculator

A sensor on a footbridge records the vertical deflection y mm of the deck as y = 5 sin x°, for 0 ≤ x ≤ 360, where x is the phase angle in degrees.

(a)

State the maximum value of y and the value of x at which it first occurs.

(2)

(b)

Explain why the equation 5 sin x° = −6 has no solutions.

(2)

(c)

Solve 5 sin x° = 3 for 0 ≤ x ≤ 360. Give your answers correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 maximum y = 5 mm	✓ 1
B1 x = 90	✓ 1
Part (b)
B1 sin x° cannot be less than −1, so the minimum of y is −5	✓ 1
B1 −6 is below −5, so no value of x works	✓ 1
Part (c)
M1 sin x° = 0.6	✓ 1
A1 x = 36.9	✓ 1
A1 x = 143.1 (180 − 36.9)	✓ 1

Common error: Giving only the calculator solution. A sine equation has a second solution in 0 to 360 found by subtracting from 180°.

+ Add to test✓ In your test289IGMC3-6-09r10-08Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium6 marks

Calculator

In triangle PQR, PQ = 14.2 km, angle QPR = 52° and angle PQR = 67°.

Diagram not to scale

(a)

Find angle PRQ.

(1)

(b)

Use the sine rule to calculate QR. Give your answer to 3 significant figures.

(3)

(c)

Calculate the area of triangle PQR. Use the unrounded value of QR and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 180° − 52° − 67° = 61°	✓ 1
Part (b)
M1 QR/sin 52° = 14.2/sin 61°	✓ 1
M1 QR = 14.2 × sin 52° ÷ sin 61°	✓ 1
A1 12.8 km	✓ 1
Part (c)
M1 area = ½ × 14.2 × QR × sin 67°	✓ 1
A1 83.6 km²	✓ 1

Exam tip: The sine rule needs a side and its opposite angle as a known pair; find the missing angle first if necessary.

+ Add to test✓ In your test290IGMC3-6-09r10-11Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium6 marks

Calculator

A barn is a cuboid ABCDEFGH. The rectangular base ABCD has AB = 20 m and BC = 15 m. The vertical edge AE = 9 m, and G is the point directly above C.

Diagram not to scale

(a)

Calculate the length of the diagonal AC of the base.

(2)

(b)

Calculate the length of AG. Give your answer to 3 significant figures.

(2)

(c)

Calculate the angle between AG and the base ABCD. Give your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 AC² = 20² + 15²	✓ 1
A1 25 m	✓ 1
Part (b)
M1 AG² = 25² + 9²	✓ 1
A1 26.6 m	✓ 1
Part (c)
M1 tan x = 9/25	✓ 1
A1 19.8°	✓ 1

Common error: For the angle between a line and a plane, use the line's projection on the plane (here AC), not an edge of the solid.

+ Add to test✓ In your test291IGMC3-6-09r11-07Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium6 marks

Calculator

A cargo ship leaves port A and sails 48 km to a point B. It then turns and sails 35 km to a point C. Angle ABC = 112°.

Diagram not to scale

(a)

Calculate the distance AC. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate angle BAC using the sine rule. Use the unrounded value of AC. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 AC² = 48² + 35² − 2 × 48 × 35 × cos 112°	✓ 1
A1 AC² = 4788 (4787.7...)	✓ 1
A1 AC = 69.2 km	✓ 1
Part (b)
M1 sin A ÷ 35 = sin 112° ÷ AC	✓ 1
M1 A = sin⁻¹(35 × sin 112° ÷ 69.19)	✓ 1
A1 28.0°	✓ 1

Common error: With an obtuse angle in the triangle, find the longest side first with the cosine rule so that the sine rule then gives an acute angle.

+ Add to test✓ In your test292IGMC3-6-09r11-10Pythagoras' theorem and right-angled trigonometry Paper EMedium8 marks

Calculator

The temperature T °C of gas leaving a fumarole t hours after midnight is modelled by T = 60 + 15 sin(30t)°, for 0 ≤ t ≤ 12, where 30t is an angle in degrees.

(a)

Write down the maximum and minimum values of T.

(2)

(b)

Write down the period of the model, in hours.

(1)

(c)

Find the two times when T = 70. Give each answer correct to 2 decimal places.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 maximum 75 °C	✓ 1
B1 minimum 45 °C	✓ 1
Part (b)
B1 360 ÷ 30 = 12 hours	✓ 1
Part (c)
M1 sin(30t) = 10 ÷ 15 = 2/3	✓ 1
A1 30t = 41.8° (41.81...)	✓ 1
A1 t = 1.39	✓ 1
M1 30t = 180° − 41.8° = 138.2°	✓ 1
A1 t = 4.61	✓ 1

Common error: A sine equation with a positive right-hand side has a second solution at 180° minus the calculator angle.

+ Add to test✓ In your test293IGMC3-6-09r11-11Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium6 marks

Calculator

A stage set is a cuboid frame ABCDEFGH. The floor ABCD is a horizontal rectangle with AB = 8.0 m and BC = 6.0 m. The vertical edges AE, BF, CG and DH are each 4.5 m high, so G is directly above C. A cable runs from A to G.

Diagram not to scale

(a)

Calculate the length of the floor diagonal AC.

(2)

(b)

Calculate the length of the cable AG. Give your answer to 3 significant figures.

(2)

(c)

Calculate the angle between the cable AG and the floor ABCD. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sqrt(8.0² + 6.0²)	✓ 1
A1 10 m	✓ 1
Part (b)
M1 sqrt(10² + 4.5²)	✓ 1
A1 11.0 m	✓ 1
Part (c)
M1 tan⁻¹(4.5 ÷ 10) (or sin⁻¹(4.5 ÷ 10.97))	✓ 1
A1 24.2°	✓ 1

Exam tip: For the angle between a line and a plane, use the right-angled triangle formed by the line, its projection on the plane and the vertical.

+ Add to test✓ In your test294IGMC3-6-08r14-10Bearings Paper EMedium6 marks

Calculator

Surveyors plan a water pipeline across a lake. They stand at A and B on the shore, with AB = 120 m. P is a point on the far shore. Angle PAB = 58° and angle PBA = 71°. Give lengths to 3 significant figures.

Diagram not to scale

(a)

Find angle APB.

(1)

(b)

Calculate the distance AP.

(3)

(c)

Calculate the shortest distance from P to the line AB. Use the unrounded value of AP.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 180° − 58° − 71° = 51°	✓ 1
Part (b)
M1 sine rule AP/sin 71° = 120/sin 51°	✓ 1
M1 AP = 120 sin 71° / sin 51°	✓ 1
A1 146 m	✓ 1
Part (c)
M1 AP × sin 58°	✓ 1
A1 124 m	✓ 1

Common error: In the sine rule each side must be paired with the angle opposite it.

+ Add to test✓ In your test295IGMC3-6-08r14-13Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium6 marks

No calculator

The concentration, C micrograms per litre, of a medicine in the blood t hours after a morning dose is modelled by C = 4.5 + 2 sin(30t)°, for 0 ≤ t ≤ 12.

(a)

Write down the maximum value of C.

(1)

(b)

Write down the minimum value of C.

(1)

(c)

Write down the period of the model, in hours.

(1)

(d)

Find the first time at which C = 5.5.

(2)

(e)

Find the second time in the interval at which C = 5.5.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 6.5	✓ 1
Part (b)
B1 2.5	✓ 1
Part (c)
B1 12	✓ 1
Part (d)
M1 2 sin(30t) = 1, so sin(30t) = 0.5	✓ 1
A1 30t = 30, t = 1	✓ 1
Part (e)
B1 30t = 150, so t = 5	✓ 1

Exam tip: For y = a + b sin(cx), the amplitude is b, the period is 360°/c and the midline is y = a.

+ Add to test✓ In your test296IGMC3-6-08r18-10Bearings Paper EMedium6 marks

Calculator

Two tracking stations A and B are on level ground, 120 km apart. They both track a relay probe P in the same vertical plane as the stations. Angle PAB = 52° and angle PBA = 71°.

Diagram not to scale

(a)

Find angle APB.

(1)

(b)

Calculate the distance AP, correct to 3 significant figures.

(3)

(c)

Calculate the distance BP, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 180° − 52° − 71° = 57°	✓ 1
Part (b)
M1 AP / sin 71° = 120 / sin 57°	✓ 1
M1 AP = 120 × sin 71° / sin 57° = 135.3	✓ 1
A1 135 km	✓ 1
Part (c)
M1 BP = 120 × sin 52° / sin 57°	✓ 1
A1 113 km	✓ 1

Exam tip: The sine rule pairs each side with the angle opposite it; find the third angle first.

+ Add to test✓ In your test297IGMC3-6-08r18-11Pythagoras' theorem and right-angled trigonometry Paper EMedium5 marks

Calculator

Two cables on a bridge meet at the top of a pylon. They are 14.2 m and 9.8 m long and the angle between them is 118°. Their lower ends are joined by a horizontal beam, forming a triangle.

Diagram not to scale

(a)

Calculate the length of the beam. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate the area of the triangle. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 beam² = 14.2² + 9.8² − 2 × 14.2 × 9.8 × cos 118°	✓ 1
M1 correct substitution giving 428.3	✓ 1
A1 20.7 m	✓ 1
Part (b)
M1 area = ½ × 14.2 × 9.8 × sin 118°	✓ 1
A1 61.4 m²	✓ 1

Common error: The cosine of an obtuse angle is negative, so the last term in the cosine rule is added.

+ Add to test✓ In your test298IGMC3-6-09r19-10Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium6 marks

Calculator

Two observers stand at points A and B on level ground, 120 m apart. They both look at the base T of a wind turbine. Angle TAB = 52° and angle TBA = 71°.

Diagram not to scale

(a)

Find angle ATB.

(1)

(b)

Use the sine rule to calculate the distance AT. Give your answer correct to 4 significant figures.

(3)

(c)

From A, the angle of elevation of the hub of the turbine is 31°. Calculate the height of the hub above the ground. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 \(180^\circ-52^\circ-71^\circ=57^\circ\)	✓ 1
Part (b)
M1 \(\frac{AT}{\sin71^\circ}=\frac{120}{\sin57^\circ}\)	✓ 1
M1 \(AT=\frac{120\sin71^\circ}{\sin57^\circ}\)	✓ 1
A1 135.3 m	✓ 1
Part (c)
M1 \(135.3\tan31^\circ\) (ECF from (b))	✓ 1
A1 81.3 m	✓ 1

Exam tip: In the sine rule each side must be paired with the angle opposite it; mark the opposite angles on your sketch first.

+ Add to test✓ In your test299IGMC3-6-09r19-11Pythagoras' theorem and right-angled trigonometry Paper EMedium7 marks

Calculator

A delivery van travels from a store S₁ to a store S₂, a distance of 8.4 km. It then travels from S₂ to a store S₃, a distance of 11.3 km. Angle S₁S₂S₃ = 108°. Use unrounded values in later parts.

Diagram not to scale

(a)

Calculate the direct distance S₁S₃. Give your answer correct to 1 decimal place.

(3)

(b)

Calculate angle S₂S₁S₃. Give your answer correct to 1 decimal place.

(2)

(c)

Calculate the area of triangle S₁S₂S₃. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(S_1S_3^2=8.4^2+11.3^2-2(8.4)(11.3)\cos108^\circ\)	✓ 1
M1 \(\sqrt{256.9}\) evaluated correctly	✓ 1
A1 16.0 km	✓ 1
Part (b)
M1 \(\frac{\sin S_2S_1S_3}{11.3}=\frac{\sin108^\circ}{16.03}\)	✓ 1
A1 42.1°	✓ 1
Part (c)
M1 \(\frac{1}{2}\times8.4\times11.3\times\sin108^\circ\)	✓ 1
A1 45.1 km\(^2\)	✓ 1

Common error: With an obtuse angle, \(\cos108^\circ\) is negative, so the term \(-2ab\cos C\) is added; entering the angle incorrectly in the calculator is the usual slip.

+ Add to test✓ In your test300IGMC3-6-09r20-06Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium6 marks

Calculator

Three beehives A, B and C stand on level ground. AB = 85 m, angle CAB = 52° and angle ABC = 71°.

Diagram not to scale

(a)

Find angle ACB.

(1)

(b)

Calculate the distance AC. Give your answer correct to 3 significant figures.

(3)

(c)

Calculate the area of triangle ABC. Use the unrounded value of AC and give your answer correct to the nearest square metre.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 57°	✓ 1
Part (b)
M1 sine rule: AC ÷ sin 71° = 85 ÷ sin 57°	✓ 1
M1 AC = 85 sin 71° ÷ sin 57°	✓ 1
A1 95.8 m	✓ 1
Part (c)
M1 area = ½ × 85 × AC × sin 52°	✓ 1
A1 3209 m²	✓ 1

Exam tip: The sine rule needs a side and its opposite angle as a known pair, so find the third angle first.

+ Add to test✓ In your test301IGMC3-6-09r20-07Pythagoras' theorem and right-angled trigonometry Paper EMedium6 marks

Calculator

Three helicopter pads P, Q and R at a hospital complex form a triangle. PQ = 7.4 km, PR = 9.1 km and angle QPR = 68°.

Diagram not to scale

(a)

Calculate the distance QR. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate angle PRQ, which is acute. Use the unrounded value of QR and give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cosine rule: QR² = 7.4² + 9.1² − 2 × 7.4 × 9.1 × cos 68°	✓ 1
M1 QR² = 87.1…	✓ 1
A1 9.33 km	✓ 1
Part (b)
M1 sine rule: sin R ÷ 7.4 = sin 68° ÷ QR	✓ 1
M1 sin R = 7.4 sin 68° ÷ QR	✓ 1
A1 47.3°	✓ 1

Exam tip: Use the cosine rule when you know two sides and the angle between them; use the sine rule next to find a missing angle.

+ Add to test✓ In your test302IGMC3-6-09r20-09Pythagoras' theorem and right-angled trigonometry Paper EMedium6 marks

No calculator

The temperature T °C inside a beehive, t hours after 06:00, is modelled by T = 33 + 4 sin(30t)°, for 0 ⩽ t ⩽ 12, where the angle 30t is in degrees.

(a)

Write down the maximum and the minimum temperature predicted by the model.

(2)

(b)

Write down the period of the model.

(1)

(c)

Find the times, in hours after 06:00, when the temperature is 35 °C.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 maximum 37 °C	✓ 1
B1 minimum 29 °C	✓ 1
Part (b)
B1 12 hours	✓ 1
Part (c)
M1 4 sin(30t) = 2 so sin(30t) = 0.5	✓ 1
A1 30t = 30 and 150	✓ 1
A1 t = 1 and t = 5	✓ 1

Exam tip: Isolate the sine term first, then use both solutions of sin x = k in the given range.

+ Add to test✓ In your test303IGMC3-6-09r21-08Pythagoras' theorem and right-angled trigonometry Paper EMedium6 marks

Calculator

A fish crate is a cuboid. Its rectangular base measures 80 cm by 60 cm and its height is 45 cm. A thin rod rests along the line from one bottom corner A to the opposite top corner G.

Diagram not to scale

(a)

Calculate the length of the diagonal of the base.

(2)

(b)

Calculate the length of the rod AG, to 3 significant figures.

(2)

(c)

Calculate the angle between the rod and the base of the crate, to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{80^2+60^2}\)	✓ 1
A1 100 cm	✓ 1
Part (b)
M1 \(\sqrt{100^2+45^2}\)	✓ 1
A1 109.66, so 110 cm	✓ 1
Part (c)
M1 \(\tan^{-1}\left(\frac{45}{100}\right)\)	✓ 1
A1 24.2°	✓ 1

Exam tip: In 3D problems, draw the right-angled triangle that contains the line and its projection onto the plane, then label it with the lengths you have found.

+ Add to test✓ In your test304IGMC3-6-09r21-09Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium6 marks

Calculator

Three marker buoys A, B and C mark the corners of a triangular fish farm. Angle BAC = 52°, angle ABC = 71° and BC = 60 m.

Diagram not to scale

(a)

Show that angle ACB = 57°.

(1)

(b)

Use the sine rule to calculate AC, to 3 significant figures.

(3)

(c)

Calculate the area of the fish farm, to 3 significant figures. Use \(AC=71.99\) m.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 \(180-52-71=57^\circ\)	✓ 1
Part (b)
M1 \(\frac{AC}{\sin71^\circ}=\frac{60}{\sin52^\circ}\)	✓ 1
M1 \(AC=\frac{60\sin71^\circ}{\sin52^\circ}\)	✓ 1
A1 72.0 m	✓ 1
Part (c)
M1 \(\frac{1}{2}\times60\times71.99\times\sin57^\circ\)	✓ 1
A1 1810 m² (ECF from (b))	✓ 1

Common error: Writing the sine rule with a side opposite the wrong angle: each side must be paired with the angle opposite it.

+ Add to test✓ In your test305IGMC3-6-09r21-10Pythagoras' theorem and right-angled trigonometry Paper EMedium5 marks

Calculator

A fashion designer cuts a triangular patch with sides of length 14 cm, 9 cm and 11 cm.

Diagram not to scale

(a)

Calculate the largest angle of the patch, to 1 decimal place.

(3)

(b)

Calculate the area of the patch, to 3 significant figures. Use the unrounded angle from part (a).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\cos\theta=\frac{9^2+11^2-14^2}{2\times9\times11}\)	✓ 1
A1 \(\cos\theta=\frac{6}{198}\), where \(\theta\) is the angle opposite the 14 cm side	✓ 1
A1 \(\theta=88.3^\circ\)	✓ 1
Part (b)
M1 \(\frac{1}{2}\times9\times11\times\sin\theta\)	✓ 1
A1 49.5 cm²	✓ 1

Exam tip: The largest angle is always opposite the longest side, so choose the cosine rule with that side as the one being 'found'.

+ Add to test✓ In your test306IGMC3-6-09r21-11Pythagoras' theorem and right-angled trigonometry Paper EMedium9 marks

Calculator

The height h metres of the tide in a fishing harbour is modelled by \(h=3+1.8\sin(30t)^\circ\), where t is the time in hours after midnight.

(a)

Write down the amplitude of the tide and the time between two successive high tides.

(2)

(b)

Find the maximum and minimum heights of the tide.

(2)

(c)

Calculate the height of the tide at 02:00, to 3 significant figures.

(2)

(d)

Find the first time after midnight when the tide height is 4.2 m. Give t to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 amplitude 1.8 m	✓ 1
B1 period \(360\div30=12\) hours	✓ 1
Part (b)
B1 maximum 4.8 m	✓ 1
B1 minimum 1.2 m	✓ 1
Part (c)
M1 \(3+1.8\sin60^\circ\)	✓ 1
A1 4.56 m	✓ 1
Part (d)
M1 \(\sin(30t)=\frac{1.2}{1.8}\)	✓ 1
M1 \(30t=\sin^{-1}\left(\frac{2}{3}\right)=41.8^\circ\)	✓ 1
A1 \(t=1.39\) hours	✓ 1

Common error: Giving the period as 30 hours. The period is the angle for one full cycle (360°) divided by the coefficient of \(t\).

+ Add to test✓ In your test307IGMC3-6-09r23-06Bearings Paper EMedium6 marks

Calculator

Two airports A and B are 340 km apart. A weather balloon is at a point P. From A the angle between the directions to B and to P is 52°, and from B the angle between the directions to A and to P is 71°.

Diagram not to scale

(a)

Find angle APB.

(1)

(b)

Use the sine rule to calculate the distance AP. Give your answer correct to 3 significant figures.

(3)

(c)

Calculate the shortest distance from P to the straight line AB. Give your answer correct to 3 significant figures. Use the unrounded value of AP.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 180 − 52 − 71 = 57°	✓ 1
Part (b)
M1 AP / sin 71° = 340 / sin 57°	✓ 1
M1 AP = 340 × sin 71° ÷ sin 57°	✓ 1
A1 383 km	✓ 1
Part (c)
M1 AP × sin 52° (accept 383.3 × sin 52°)	✓ 1
A1 302 km	✓ 1

Exam tip: Find the third angle first so that you can pair every side with its opposite angle.

+ Add to test✓ In your test308IGMC3-6-09r23-11Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium6 marks

Calculator

A grain store is a cuboid with a horizontal rectangular base ABCD and top EFGH, where E is above A, F above B, G above C and H above D. AB = 8 m, BC = 6 m and the height AE = 5 m.

Diagram not to scale

(a)

Calculate the length of the diagonal AC of the base.

(2)

(b)

Calculate the length of the space diagonal AG. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the angle between AG and the base ABCD. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sqrt(8² + 6²)	✓ 1
A1 10 m	✓ 1
Part (b)
M1 sqrt(10² + 5²)	✓ 1
A1 11.2 m	✓ 1
Part (c)
M1 tan x = 5/10	✓ 1
A1 26.6°	✓ 1

Exam tip: For angles in 3D, find the right-angled triangle that contains the line and its projection on the plane.

+ Add to test✓ In your test309IGMC3-6-10r25-11Bearings Paper EMedium6 marks

Calculator

Two delivery vans leave a warehouse \(W\) at the same time. Van 1 drives 18 km on a bearing of 035°. Van 2 drives 25 km on a bearing of 118°. Van 1 is then at \(P\) and van 2 is at \(R\).

Diagram not to scale

(a)

Find angle \(PWR\).

(1)

(b)

Calculate the distance \(PR\), to 3 significant figures.

(3)

(c)

Calculate the area of triangle \(WPR\), to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 \(118° - 35° = 83°\)	✓ 1
Part (b)
M1 cosine rule \(PR^2 = 18^2 + 25^2 - 2(18)(25)\cos 83°\)	✓ 1
A1 \(PR^2 = 839\) (or 839.3)	✓ 1
A1 29.0 km	✓ 1
Part (c)
M1 \(\frac{1}{2} \times 18 \times 25 \times \sin 83°\)	✓ 1
A1 223 km²	✓ 1

Common error: In the cosine rule, evaluate \(2ab\cos C\) first and then subtract it; subtracting before multiplying by the cosine gives a wrong result.

+ Add to test✓ In your test310IGMC3-6-10r25-12Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium8 marks

Calculator

The depth of water \(h\) metres at a marine research jetty is modelled by \(h = 3.2 + 1.8\sin(30t)°\), where \(t\) is the time in hours after midnight and \(0 \le t \le 12\).

(a)

State the maximum and minimum depths of water predicted by the model.

(2)

(b)

Calculate the depth at \(t = 2\), to 3 significant figures.

(2)

(c)

State the time, in hours, between one high water and the next.

(1)

(d)

Solve \(3.2 + 1.8\sin(30t)° = 4.1\) for \(0 \le t \le 12\).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 maximum 5.0 m	✓ 1
B1 minimum 1.4 m	✓ 1
Part (b)
M1 \(3.2 + 1.8\sin 60°\)	✓ 1
A1 4.76 m	✓ 1
Part (c)
B1 period \(= \frac{360}{30} = 12\) hours	✓ 1
Part (d)
M1 \(\sin(30t)° = 0.5\) so \(30t = 30\)	✓ 1
A1 \(t = 1\)	✓ 1
A1 \(t = 5\) (from \(30t = 150\))	✓ 1

Exam tip: A sine curve has two solutions in each cycle for most heights: use the symmetry \(180° - \theta\) to find the second solution.

+ Add to test✓ In your test311IGMC3-6-10r25-13Pythagoras' theorem and right-angled trigonometry Paper EMedium6 marks

Calculator

A rectangular atrium has a horizontal floor \(ABCD\) that is 8 m long (\(AB\)) and 6 m wide (\(BC\)). The vertical height of the atrium is 3.5 m. A support cable runs in a straight line from floor corner \(A\) to the ceiling corner directly above \(C\), called \(G\).

Diagram not to scale

(a)

Calculate the length of the floor diagonal \(AC\).

(2)

(b)

Calculate the length of cable \(AG\), to 3 significant figures.

(2)

(c)

Calculate the angle between the cable \(AG\) and the floor, to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{8^2 + 6^2}\)	✓ 1
A1 10 m	✓ 1
Part (b)
M1 \(\sqrt{10^2 + 3.5^2}\) (ECF from (a))	✓ 1
A1 10.6 m	✓ 1
Part (c)
M1 \(\tan^{-1}\left(\frac{3.5}{10}\right)\) (or \(\sin^{-1}\left(\frac{3.5}{AG}\right)\))	✓ 1
A1 19.3°	✓ 1

Common error: The angle with the floor is found in the vertical triangle \(ACG\) using \(AC\), not \(AB\) or \(BC\).

+ Add to test✓ In your test312IGMC3-6-08r26-07Pythagoras' theorem and right-angled trigonometry Paper EMedium6 marks

No calculator

A brace on a bridge forms a right-angled triangle ABC. Angle ABC = 90°, angle BAC = 60° and AB = 8 m. Give all of your answers in exact form.

Diagram not to scale

(a)

Find the length of BC.

(2)

(b)

Find the length of AC.

(2)

(c)

Find the area of triangle ABC.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 BC = 8 tan 60°	✓ 1
A1 8√3 m	✓ 1
Part (b)
M1 cos 60° = 8 ÷ AC	✓ 1
A1 AC = 16 m	✓ 1
Part (c)
M1 ½ × 8 × 8√3	✓ 1
A1 32√3 m² (= 55.4 m² to 3 significant figures)	✓ 1

Exam tip: Learn the exact values: tan 60° = √3, cos 60° = 1/2, sin 60° = √3/2, so that you can work without a calculator.

+ Add to test✓ In your test313IGMC3-6-08r26-11Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium7 marks

No calculator

During a roasting cycle, the temperature of the air in a coffee roasting drum is T °C at t minutes after the start of the cycle, where T = 210 + 12 sin(45t)° for 0 ≤ t ≤ 8.

(a)

Write down the maximum temperature and the minimum temperature of the air in the drum.

(2)

(b)

Write down the period of the function, in minutes.

(1)

(c)

Find the two times at which the temperature of the air is 216 °C. Give your answers as exact fractions.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 maximum 222 °C	✓ 1
B1 minimum 198 °C	✓ 1
Part (b)
B1 360 ÷ 45 = 8 minutes	✓ 1
Part (c)
M1 12 sin(45t) = 6, so sin(45t) = ½	✓ 1
A1 45t = 30, so t = 2/3 minutes	✓ 1
M1 45t = 180 − 30 = 150	✓ 1
A1 t = 10/3 minutes (= 3.33)	✓ 1

Common error: A sine equation usually has a second solution in each cycle: use 180° minus the first angle.

+ Add to test✓ In your test314IGMC3-6-11r27-09Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium6 marks

Calculator

Two microphones M and N stand on level ground 6.0 m apart. A singer S stands so that angle SMN = 52° and angle SNM = 71°.

Diagram not to scale

(a)

Find angle MSN.

(1)

(b)

Calculate the distance MS. Give your answer to 3 significant figures.

(3)

(c)

Calculate the shortest distance from S to the line MN. Use the unrounded value of MS. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 180° − 52° − 71° = 57°	✓ 1
Part (b)
M1 sine rule: MS ÷ sin 71° = 6.0 ÷ sin 57°	✓ 1
M1 MS = 6.0 × sin 71° ÷ sin 57°	✓ 1
A1 6.76 m	✓ 1
Part (c)
M1 distance = MS × sin 52°	✓ 1
A1 5.33 m	✓ 1

Common error: In the sine rule each side must be paired with the angle opposite it: MS is opposite angle N (71°), not angle M.

+ Add to test✓ In your test315IGMC3-6-10r30-10Bearings Paper EMedium6 marks

Calculator

Two surveyors stand at points A and B on one bank of a river, 120 m apart. They look at the top P of a bridge pylon on the other bank. Angle PAB = 54° and angle PBA = 71°.

(a)

Find angle APB.

(1)

(b)

Calculate the distance AP. Give your answer correct to 3 significant figures.

(3)

(c)

Calculate the shortest distance from P to the line AB. Use the unrounded value of AP and give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 \(180-54-71=55°\)	✓ 1
Part (b)
M1 sine rule \(\frac{AP}{\sin 71°}=\frac{120}{\sin 55°}\)	✓ 1
M1 \(AP=\frac{120\sin 71°}{\sin 55°}\)	✓ 1
A1 139 m	✓ 1
Part (c)
M1 \(AP\sin 54°\)	✓ 1
A1 112 m	✓ 1

Exam tip: Use the sine rule when you know a side and its opposite angle, and find the missing angle first if needed.

+ Add to test✓ In your test316IGMC3-6-10r30-11Pythagoras' theorem and right-angled trigonometry Paper EMedium5 marks

Calculator

Two tracking dishes, D1 and D2, are 7.2 km and 9.5 km from a control tower T. The angle D1TD2 is 64°.

(a)

Calculate the distance D1D2. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate the area of triangle D1TD2. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cosine rule \(D_1D_2^2=7.2^2+9.5^2-2(7.2)(9.5)\cos 64°\)	✓ 1
A1 \(D_1D_2^2=82.1...\)	✓ 1
A1 9.06 km	✓ 1
Part (b)
M1 \(\frac{1}{2}(7.2)(9.5)\sin 64°\)	✓ 1
A1 30.7 km²	✓ 1

Common error: In the cosine rule the angle must be the one between the two known sides.

+ Add to test✓ In your test317IGMC3-6-11r33-06Bearings Paper EMedium5 marks

Calculator

In a strategy game, a player's base B is 80 m from a tower P and 125 m from a tower Q. The angle PBQ is 64°.

(a)

Calculate the distance PQ. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate the area of triangle PBQ. Give your answer correct to the nearest square metre.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 PQ² = 80² + 125² − 2 × 80 × 125 × cos 64°	✓ 1
M1 PQ² = 13257.6... (accept 13 300)	✓ 1
A1 115 m	✓ 1
Part (b)
M1 ½ × 80 × 125 × sin 64°	✓ 1
A1 4494 m²	✓ 1

Exam tip: Use the cosine rule when you know two sides and the angle between them; use ½ab sin C for the area with the same data.

+ Add to test✓ In your test318IGMC3-6-11r33-07Bearings Paper EMedium8 marks

Calculator

Two pumping stations A and B lie on a straight, level road running east to west, with B 2.4 km due east of A. A reservoir R lies north of the road. The bearing of R from A is 055° and the bearing of R from B is 312°.

(a)

Find angle RAB, angle RBA and angle ARB.

(3)

(b)

Calculate the distance AR. Give your answer correct to 3 significant figures. Keep the unrounded value for part (c).

(3)

(c)

Calculate the shortest distance from R to the road. Give your answer correct to 2 decimal places.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 RAB = 90° − 55° = 35°	✓ 1
B1 RBA = 312° − 270° = 42°	✓ 1
B1 ARB = 180° − 35° − 42° = 103°	✓ 1
Part (b)
M1 AR/sin 42° = 2.4/sin 103°	✓ 1
M1 AR = 2.4 sin 42°/sin 103°	✓ 1
A1 1.65 km	✓ 1
Part (c)
M1 AR × sin 35° (or 2.4 sin 35° sin 42°/sin 103°)	✓ 1
A1 0.95 km	✓ 1

Common error: Draw a north line at each point before reading off angles from bearings, because the angle inside the triangle is not the bearing itself.

+ Add to test✓ In your test319IGMC3-6-11r33-08Pythagoras' theorem and right-angled trigonometry Paper EMedium8 marks

Calculator

The depth, d metres, of water in a harbour t hours after midnight is modelled by d = 6 + 2.5 sin(30t)°, for 0 ≤ t ≤ 12, where the angle 30t is measured in degrees.

(a)

Write down the maximum depth and the minimum depth of the water.

(2)

(b)

Write down the period of the model.

(1)

(c)

Calculate the depth of water at 02:00. Give your answer correct to 3 significant figures.

(2)

(d)

Find the first time after midnight at which the depth is 7 m. Give your answer in hours correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 maximum 8.5 m	✓ 1
B1 minimum 3.5 m	✓ 1
Part (b)
B1 12 hours	✓ 1
Part (c)
M1 6 + 2.5 sin 60°	✓ 1
A1 8.17 m	✓ 1
Part (d)
M1 sin(30t) = 0.4	✓ 1
M1 30t = sin⁻¹ 0.4 = 23.58...°	✓ 1
A1 t = 0.786 hours	✓ 1

Exam tip: In a sine model of the form a + b sin(ct), a is the midline and b is the amplitude, so the extremes are a ± b.

+ Add to test✓ In your test320IGMC3-6-10r38-08Bearings Paper EMedium6 marks

Calculator

Two research stations A and B are 6.4 km apart on a straight, level stretch of coastline. A whale W is seen in the sea. Angle WAB = 58° and angle WBA = 71°, and W is on the sea side of the coastline. Give answers to 3 significant figures.

A and B are on the coastline; W is out at sea. Diagram not to scale

(a)

Find angle AWB.

(1)

(b)

Calculate the distance AW.

(3)

(c)

Calculate the shortest distance from the whale to the coastline. Use the unrounded value of AW.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 180 − 58 − 71 = 51°	✓ 1
Part (b)
M1 AW ÷ sin 71° = 6.4 ÷ sin 51° (sine rule)	✓ 1
M1 AW = 6.4 × sin 71° ÷ sin 51°	✓ 1
A1 7.79 km	✓ 1
Part (c)
M1 AW × sin 58°	✓ 1
A1 6.60 km	✓ 1

Exam tip: With two angles and one side, find the third angle first so that the sine rule can be set up with a complete side-angle pair.

+ Add to test✓ In your test321IGMC3-6-10r38-10Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium7 marks

Calculator

A triangular glass panel in a building has two sides of length 4.6 m and 5.3 m with an angle of 68° between them.

Diagram not to scale

(a)

Calculate the area of the panel, to 3 significant figures.

(2)

(b)

Calculate the length of the third side of the panel, to 3 significant figures.

(3)

(c)

Glass costs £85 per m². Use the unrounded area from part (a) to calculate the cost of the glass for one panel, to the nearest pound.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 ½ × 4.6 × 5.3 × sin 68°	✓ 1
A1 11.3 m² (11.302... before rounding)	✓ 1
Part (b)
M1 4.6² + 5.3² − 2 × 4.6 × 5.3 × cos 68° (cosine rule)	✓ 1
M1 square root of 30.98...	✓ 1
A1 5.57 m	✓ 1
Part (c)
M1 11.302... × 85	✓ 1
A1 £961	✓ 1

Exam tip: Use the area formula ½ab sin C when the angle given lies between the two known sides.

+ Add to test✓ In your test322IGMC3-6-10r38-11Pythagoras' theorem and right-angled trigonometry Paper EMedium6 marks

Calculator

A storage container in the shape of a cuboid has length 12 m, width 9 m and height 8 m. A cable is stretched from one bottom corner A of the container, through the inside, to the opposite top corner G, which is diagonally opposite A.

Diagram not to scale

(a)

Calculate the length of the diagonal of the base of the container.

(2)

(b)

Calculate the length AG of the cable.

(2)

(c)

Calculate the angle between the cable AG and the base of the container, giving your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12² + 9² (= 225)	✓ 1
A1 15 m	✓ 1
Part (b)
M1 15² + 8² (= 289), ECF from (a)	✓ 1
A1 17 m	✓ 1
Part (c)
M1 tan θ = 8 ÷ 15 (or sin θ = 8 ÷ 17, or cos θ = 15 ÷ 17)	✓ 1
A1 28.1°	✓ 1

Common error: The angle between a line and a plane is measured to its projection on the plane, here the base diagonal, not to an edge of the base.

+ Add to test✓ In your test323IGMC3-6-10r40-10Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium6 marks

Calculator

Two telecoms towers A and B are on level ground 9.0 km apart. A signal source C is such that angle CAB = 52° and angle ABC = 71°.

Diagram not to scale

(a)

Find angle ACB.

(1)

(b)

Use the sine rule to calculate the distance AC. Give your answer to 3 significant figures.

(3)

(c)

Calculate the distance BC. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 57°	✓ 1
Part (b)
M1 \(\dfrac{AC}{\sin 71°}=\dfrac{9.0}{\sin 57°}\)	✓ 1
M1 \(AC=\dfrac{9.0\sin 71°}{\sin 57°}\)	✓ 1
A1 10.1 km	✓ 1
Part (c)
M1 \(BC=\dfrac{9.0\sin 52°}{\sin 57°}\)	✓ 1
A1 8.46 km	✓ 1

Exam tip: The sine rule pairs each side with the angle opposite it, so first find the angle opposite the known side.

+ Add to test✓ In your test324IGMC3-6-10r40-11Pythagoras' theorem and right-angled trigonometry Paper EMedium6 marks

Calculator

Two climbers leave the same base camp at the same time. Climber 1 walks 6.5 km on a bearing of 038°. Climber 2 walks 4.2 km on a bearing of 112°. Both walk in straight lines on level ground.

O is the base camp. Diagram not to scale

(a)

Show that the angle between their paths is 74°.

(1)

(b)

Calculate the distance between the climbers when they stop. Give your answer to 3 significant figures.

(3)

(c)

Calculate the area of the triangle formed by the base camp and the two climbers. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 \(112-38=74\)	✓ 1
Part (b)
M1 \(d^2=6.5^2+4.2^2-2(6.5)(4.2)\cos 74°\)	✓ 1
A1 \(d^2=44.84\) (accept 44.8)	✓ 1
A1 6.70 km	✓ 1
Part (c)
M1 \(\dfrac12\times 6.5\times 4.2\sin 74°\)	✓ 1
A1 13.1 km²	✓ 1

Common error: In the cosine rule the angle must be the one between the two known sides, and the 2ab cos C term is subtracted.

+ Add to test✓ In your test325IGMC3-6-10r40-12Bearings Paper EMedium6 marks

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The temperature, \(T\) °C, inside a bread oven during a baking cycle is modelled by \(T=180+12\sin(30t)\), where \(t\) is the time in minutes after the cycle starts and \(30t\) is in degrees.

(a)

Write down the maximum and minimum temperatures predicted by the model.

(2)

(b)

Find the time taken for one complete temperature cycle.

(1)

(c)

Find the first time at which the temperature reaches 189 °C. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 192 °C	✓ 1
B1 168 °C	✓ 1
Part (b)
B1 12 minutes	✓ 1
Part (c)
M1 \(12\sin(30t)=9\) or \(\sin(30t)=0.75\)	✓ 1
M1 \(30t=\sin^{-1}0.75=48.59°\)	✓ 1
A1 1.62 minutes	✓ 1

Exam tip: For the first time, take the smallest positive angle from the inverse sine; later times come from using 180° minus that angle and adding full cycles.

+ Add to test✓ In your test326IGMC3-6-11r40-12Bearings Paper EMedium7 marks

No calculator

The temperature, T °C, of a coffee-roasting drum during one cycle is modelled by T = 205 + 12 sin(30t)°, where t is the time in minutes and 0 ≤ t ≤ 12.

(a)

Write down the period of the function, in minutes.

(1)

(b)

Write down the maximum and the minimum temperature during the cycle.

(2)

(c)

Find the first time at which the temperature is 211 °C.

(2)

(d)

Find the second time in the cycle at which the temperature is 211 °C.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 12	✓ 1
Part (b)
B1 217 °C	✓ 1
B1 193 °C	✓ 1
Part (c)
M1 \(12\sin(30t)=6\) so \(\sin(30t)=0.5\)	✓ 1
A1 \(t=1\) minute (from \(30t=30\))	✓ 1
Part (d)
M1 \(30t=180-30=150\)	✓ 1
A1 \(t=5\) minutes	✓ 1

Common error: Stopping at the first solution from the calculator or exact values. The sine curve takes each value between its maximum and minimum twice in every period, so look for 180° minus the first angle.

+ Add to test✓ In your test327IGMC3-6-10r45-10Pythagoras' theorem and right-angled trigonometry Paper EMedium8 marks

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An archaeological site is in the shape of triangle ABC with AB = 14.0 m, AC = 9.5 m and angle BAC = 62°.

Diagram not to scale

(a)

Calculate the length of BC. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate angle ABC. Give your answer correct to 1 decimal place.

(3)

(c)

Calculate the area of the site. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 BC² = 14.0² + 9.5² − 2 × 14.0 × 9.5 × cos 62°	✓ 1
A1 BC² = 161.4 (161.37...)	✓ 1
A1 BC = 12.7 m	✓ 1
Part (b)
M1 sin B ÷ 9.5 = sin 62° ÷ BC	✓ 1
A1 sin B = 0.660	✓ 1
A1 41.3° (accept answers using 12.7 or the unrounded value of BC)	✓ 1
Part (c)
M1 ½ × 14.0 × 9.5 × sin 62°	✓ 1
A1 58.7 m²	✓ 1

Exam tip: Use the cosine rule when you know two sides and the angle between them; then switch to the sine rule for another angle.

+ Add to test✓ In your test328IGMC3-6-10r45-11Bearings Paper EMedium9 marks

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Two mountain huts A and B are 6.4 km apart, with B due east of A. A walker is at point C. From A, the bearing of C is 050°. From B, the bearing of C is 335°.

Diagram not to scale

(a)

Show that angle ACB = 75°.

(3)

(b)

Calculate the distance AC. Give your answer correct to 3 significant figures.

(3)

(c)

Write down the bearing of A from C.

(1)

(d)

Calculate the distance BC. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 angle CAB = 090° − 050° = 40°	✓ 1
B1 angle ABC = 335° − 270° = 65°	✓ 1
A1 180° − 40° − 65° = 75°	✓ 1
Part (b)
M1 AC ÷ sin 65° = 6.4 ÷ sin 75°	✓ 1
M1 AC = 6.4 × sin 65° ÷ sin 75°	✓ 1
A1 6.00 km	✓ 1
Part (c)
B1 230°	✓ 1
Part (d)
M1 BC = 6.4 × sin 40° ÷ sin 75°	✓ 1
A1 4.26 km	✓ 1

Exam tip: In the sine rule, each side pairs with the angle opposite it, so find the third angle first.

+ Add to test✓ In your test329IGMC3-6-12r45-07Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium6 marks

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At a Formula 1 circuit three marshal posts A, B and C form a triangle. AB = 420 m, angle CAB = 52° and angle ABC = 71°.

Diagram not to scale

(a)

Find angle ACB.

(1)

(b)

Calculate the distance BC. Give your answer to 3 significant figures.

(3)

(c)

Calculate the shortest distance from C to the line AB. Use the unrounded value of BC and give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
180° − 52° − 71° = 57°	✓ 1
Part (b)
M1 BC ÷ sin 52° = 420 ÷ sin 57°	✓ 1
M1 BC = 420 × sin 52° ÷ sin 57°	✓ 1
A1 395 m	✓ 1
Part (c)
M1 distance = BC × sin 71°	✓ 1
A1 373 m	✓ 1

Exam tip: Use the sine rule when you know a side and its opposite angle; find the third angle first if needed.

+ Add to test✓ In your test330IGMC3-6-12r45-08Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium5 marks

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A triangular plot of land has sides of length 7.8 m, 9.1 m and 12.4 m. Hives are to be placed inside it.

Diagram not to scale

(a)

Calculate the largest angle of the plot. Give your answer to 1 decimal place.

(3)

(b)

Calculate the area of the plot. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cos C = (7.8² + 9.1² − 12.4²) ÷ (2 × 7.8 × 9.1)	✓ 1
A1 cos C = −0.0712 (negative, so the angle is obtuse)	✓ 1
A1 94.1°	✓ 1
Part (b)
M1 ½ × 7.8 × 9.1 × sin 94.08°	✓ 1
A1 35.4 m²	✓ 1

Exam tip: The largest angle is opposite the longest side; a negative cosine means the angle is obtuse.

+ Add to test✓ In your test331IGMC3-6-12r45-10Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium6 marks

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A hospital storage room is a cuboid of length 8.0 m, width 5.0 m and height 3.2 m. A straight cable runs from one bottom corner to the opposite top corner.

Diagram not to scale

(a)

Calculate the length of the diagonal of the floor. Give your answer to 3 significant figures.

(2)

(b)

Calculate the length of the cable. Use the unrounded value from part (a) and give your answer to 3 significant figures.

(2)

(c)

Calculate the angle between the cable and the floor. Give your answer to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(8.0² + 5.0²)	✓ 1
A1 9.43 m	✓ 1
Part (b)
M1 √(89 + 3.2²)	✓ 1
A1 9.96 m	✓ 1
Part (c)
M1 tan θ = 3.2 ÷ 9.43 (or sin θ = 3.2 ÷ 9.96)	✓ 1
A1 18.7°	✓ 1

Exam tip: In 3D problems, find a right-angled triangle first (here the floor diagonal and the height) and then apply Pythagoras or trigonometry.

+ Add to test✓ In your test332IGMC3-6-12r45-12Pythagoras' theorem and right-angled trigonometry Paper EMedium7 marks

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The temperature, T °C, inside a beehive t hours after midnight is modelled by
T = 18 + 6 sin(15t°), for 0 ≤ t ≤ 24.

(a)

Write down the amplitude and the period of this function.

(2)

(b)

Write down the maximum and minimum temperatures.

(2)

(c)

Find the times when the temperature is 21 °C.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 amplitude 6	✓ 1
B1 period 360 ÷ 15 = 24 hours	✓ 1
Part (b)
B1 maximum 24 °C	✓ 1
B1 minimum 12 °C	✓ 1
Part (c)
M1 sin(15t°) = 0.5	✓ 1
A1 15t = 30, so t = 2	✓ 1
A1 15t = 150, so t = 10	✓ 1

Common error: A sine equation usually has a second solution in the range: for sin x = 0.5, use 180° − 30° = 150° as well as 30°.

+ Add to test✓ In your test333IGMC3-6-11r46-04Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EMedium7 marks

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Two observation posts, A and B, are on a straight shoreline and are 1.8 km apart. A whale W is in the sea. The angle WAB is 52° and the angle ABW is 71°.

A and B are on the shoreline; W is out at sea. Diagram not to scale

(a)

Show that angle AWB is 57°.

(2)

(b)

Calculate the distance AW. Give your answer to 3 significant figures.

(3)

(c)

Calculate the shortest distance from the whale to the shoreline. Give your answer to 3 significant figures. Use the unrounded value of AW.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses angle sum of a triangle, \(180-52-71\)	✓ 1
A1 57° stated (shown)	✓ 1
Part (b)
M1 sine rule with side AB opposite the 57° angle	✓ 1
M1 \(AW=\frac{1.8\sin71^\circ}{\sin57^\circ}\)	✓ 1
A1 2.03 km	✓ 1
Part (c)
M1 \(AW\sin52^\circ\) (or \(1.8\sin71^\circ\sin52^\circ\div\sin57^\circ\))	✓ 1
A1 1.60 km	✓ 1

Exam tip: In the sine rule, each side must be paired with the angle opposite it, so find the missing angle first if needed.

+ Add to test✓ In your test334IGMC3-6-11r46-06Bearings Paper EMedium6 marks

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Two delivery vans leave a depot D at the same time on different straight roads. After one hour, van 1 is 24 km from D and van 2 is 31 km from D. The angle between the two roads is 68°.

Diagram not to scale

(a)

Calculate the distance between the two vans after one hour. Give your answer to 3 significant figures.

(4)

(b)

A drone flies directly from van 1 to van 2 at a constant speed of 150 km/h. Calculate the time taken in minutes, to 1 decimal place. Use the unrounded distance from (a).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cosine rule with the 68° angle between sides 24 and 31	✓ 1
M1 \(24^2+31^2-2\times24\times31\cos68^\circ\)	✓ 1
M1 square root of the correct expression	✓ 1
A1 31.3 km	✓ 1
Part (b)
M1 \(\frac{31.298}{150}\times60\)	✓ 1
A1 12.5 minutes	✓ 1

Exam tip: The cosine rule for a side needs the angle between the two known sides; check that your calculator is in degree mode.

+ Add to test✓ In your test335IGMC3-6-11r46-11Pythagoras' theorem and right-angled trigonometry Paper EMedium8 marks

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The height, h metres, of the tide in a harbour is modelled by h = 2.4 + 1.8cos(30t)°, where t is the time in hours after midnight.

(a)

Write down the maximum and minimum heights of the tide.

(2)

(b)

Write down the time between two consecutive high tides.

(1)

(c)

Calculate the height of the tide at 02:00.

(2)

(d)

Calculate the first time after midnight when the height of the tide is 3.0 m. Give your answer in hours to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 4.2 m	✓ 1
B1 0.6 m	✓ 1
Part (b)
B1 12 hours	✓ 1
Part (c)
M1 substitutes \(t=2\), so \(\cos60^\circ\)	✓ 1
A1 3.3 m	✓ 1
Part (d)
M1 \(\cos(30t)=\frac{0.6}{1.8}\)	✓ 1
M1 inverse cosine, \(30t=70.5^\circ\)	✓ 1
A1 2.35 hours	✓ 1

Exam tip: For the first solution after t = 0, take the principal value from the inverse cosine, then divide by the coefficient of t.

+ Add to test✓ In your test336IGMC-046-01-12Bearings Paper EHard7 marks

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A lighthouse L is 30 km due east of a port P. A ship leaves P and sails in a straight line on a bearing of 050° at a constant speed of 18 km/h.

Diagram not to scale

(a)

Calculate the shortest distance between the ship's path and the lighthouse.

(2)

(b)

Calculate how long after leaving P the ship is closest to the lighthouse. Give your answer in hours and minutes, to the nearest minute.

(3)

(c)

Find the bearing of the lighthouse from the ship at the moment it is closest.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 30 × sin 40°, using angle between path and PL = 90 − 50	✓ 1
A1: 19.3 km	✓ 1
Part (b)
M1: distance along path = 30 × cos 40° (= 22.98 km)	✓ 1
M1: their distance ÷ 18 (= 1.277 h)	✓ 1
A1: 1 hour 17 minutes	✓ 1
Part (c)
M1: path is perpendicular to the line from ship to L, so 50 + 90	✓ 1
A1: 140°	✓ 1

Exam tip: The shortest distance from a point to a straight line is always measured along the perpendicular. This creates a right angle at the ship's closest position.

+ Add to test✓ In your test337IGMC-046-01-13Bearings Paper EHard8 marks

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A boat leaves a jetty A and sails 14 km on a bearing of 125° to a buoy B. An island C is 20 km from A on a bearing of 070°.

Diagram not to scale

(a)

Calculate the distance BC.

(3)

(b)

Calculate the area of triangle ABC.

(2)

(c)

Calculate the bearing of C from B.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: angle BAC = 125 − 70 = 55°	✓ 1
M1: BC² = 14² + 20² − 2 × 14 × 20 × cos 55°	✓ 1
A1: 16.6 km	✓ 1
Part (b)
M1: ½ × 14 × 20 × sin 55°	✓ 1
A1: 115 km²	✓ 1
Part (c)
M1: cos ABC = (14² + BC² − 20²) ÷ (2 × 14 × BC), or a correct sine rule with a check that the angle is acute	✓ 1
A1: angle ABC = 81.2°	✓ 1
A1: bearing = 305 + 81.2 − 360 = 026.2°	✓ 1

Common error: The sine rule gives sin ABC ≈ 0.988, which matches both 81.2° and 98.8°. Using the cosine rule for the angle avoids choosing the wrong one.

+ Add to test✓ In your test338IGMC-046-01-14Bearings Paper EHard8 marks

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Three towns X, Y and Z are such that XY = 45 km, YZ = 38 km and XZ = 52 km. The bearing of Y from X is 080°. Z lies to the south of the line XY.

Diagram not to scale

(a)

Calculate angle YXZ.

(3)

(b)

Find the bearing of Z from X.

(1)

(c)

Calculate the bearing of Z from Y.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: cos YXZ = (45² + 52² − 38²) ÷ (2 × 45 × 52)	✓ 1
A1: = 3285 ÷ 4680 = 0.7019…	✓ 1
A1: 45.4°	✓ 1
Part (b)
B1: 80 + 45.4 = 125.4°	✓ 1
Part (c)
M1: cos XYZ = (45² + 38² − 52²) ÷ (2 × 45 × 38)	✓ 1
A1: angle XYZ = 77.1°	✓ 1
M1: bearing of X from Y = 260°, then 260 − their angle XYZ	✓ 1
A1: 182.9°	✓ 1

Exam tip: Decide which side of the line the third point lies on before you add or subtract. Z is south of XY, so from X it is clockwise of Y, and from Y it is anticlockwise of X.

+ Add to test✓ In your test339IGMC-046-00-12Pythagoras' theorem and right-angled trigonometry Paper EHard7 marks

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A cuboid ABCDEFGH has a rectangular base ABCD. AB = 12 cm, BC = 5 cm and the vertical edge CG = 4 cm. G is vertically above C.

Diagram not to scale

(a)

Calculate the length of the diagonal AC of the base.

(2)

(b)

Calculate the length of AG.

(2)

(c)

Calculate the angle between AG and the base ABCD.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: √(12² + 5²)	✓ 1
A1: 13 cm	✓ 1
Part (b)
M1: √(13² + 4²) or √(12² + 5² + 4²)	✓ 1
A1: 13.6 cm (13.60…)	✓ 1
Part (c)
B1: angle GAC identified as the required angle	✓ 1
M1: tan θ = 4 ÷ 13 (or equivalent using sin or cos)	✓ 1
A1: 17.1°	✓ 1

Exam tip: The angle between a line and a plane is the angle between the line and its projection onto the plane. Here the projection of AG is AC, so work in triangle ACG.

+ Add to test✓ In your test340IGMC-046-00-13Pythagoras' theorem and right-angled trigonometry Paper EHard7 marks

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A right-angled triangle has sides of length x cm, (x + 7) cm and (x + 8) cm, where (x + 8) cm is the hypotenuse.

Diagram not to scale

(a)

Show that x² − 2x − 15 = 0.

(3)

(b)

Solve the equation to find the value of x.

(2)

(c)

Calculate the smallest angle in the triangle.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: x² + (x + 7)² = (x + 8)²	✓ 1
M1: expands correctly: 2x² + 14x + 49 = x² + 16x + 64	✓ 1
A1: collects terms to reach x² − 2x − 15 = 0 with no errors	✓ 1
Part (b)
M1: (x − 5)(x + 3) = 0, or correct substitution into the quadratic formula	✓ 1
A1: x = 5 only, with −3 rejected because a length must be positive	✓ 1
Part (c)
M1: a correct trigonometric ratio for the angle opposite the 5 cm side, e.g. tan θ = 5 ÷ 12, sin θ = 5 ÷ 13 or cos θ = 12 ÷ 13	✓ 1
A1: 22.6°	✓ 1

Common error: (x + 7)² is not x² + 49. Expand it fully as x² + 14x + 49.

+ Add to test✓ In your test341IGMC-046-00-14Pythagoras' theorem and right-angled trigonometry Paper EHard5 marks

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A vertical mast AB stands with its base B on level ground. Points C and D are on the ground in a straight line with B, on the same side of the mast. D is further from the mast than C, and CD = 20 m.
The angle of elevation of A from C is 41°. The angle of elevation of A from D is 27°.
The height of the mast is h metres.

Diagram not to scale

(a)

Write an expression for BC in terms of h.

(1)

(b)

Write an expression for BD in terms of h.

(1)

(c)

Calculate h.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1: BC = h ÷ tan 41°	✓ 1
Part (b)
B1: BD = h ÷ tan 27°	✓ 1
Part (c)
M1: h ÷ tan 27° − h ÷ tan 41° = 20	✓ 1
M1: h = 20 ÷ (1 ÷ tan 27° − 1 ÷ tan 41°)	✓ 1
A1: 24.6 m (24.62…)	✓ 1

Exam tip: Neither triangle has enough information on its own. Write both horizontal distances in terms of h, then use the fact that they differ by 20 m.

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ABCD is a quadrilateral field. The diagonal AC divides it into triangles ABC and ACD.
AB = 6 m, BC = 9 m and angle ABC = 104°.
CD = 7.5 m and angle ACD = 38°.

Diagram not to scale

(a)

Calculate the length of AC.

(3)

(b)

Calculate the length of AD.

(3)

(c)

Calculate the total area of the field ABCD.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: AC² = 6² + 9² − 2 × 6 × 9 × cos 104°	✓ 1
A1: AC² = 143.1(3…)	✓ 1
A1: AC = 11.96 m (11.963…)	✓ 1
Part (b)
M1: AD² = 11.96² + 7.5² − 2 × 11.96 × 7.5 × cos 38°	✓ 1
A1: AD² = 57.9(7…)	✓ 1
A1: AD = 7.61 m	✓ 1
Part (c)
M1: area ABC = ½ × 6 × 9 × sin 104° (= 26.19…)	✓ 1
M1: area ACD = ½ × their 11.96 × 7.5 × sin 38° (= 27.62…)	✓ 1
A1: total area = 53.8 m²	✓ 1

Exam tip: Store AC in your calculator memory instead of re-entering a rounded value. Both later parts depend on it, and early rounding can change the third significant figure.

+ Add to test✓ In your test343IGMC-046-02-13Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EHard8 marks

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A vertical mast TF is 40 m tall, with its base F on horizontal ground. Point A is due south of F, and the angle of elevation of T from A is 25°. Point B is due east of F, and the angle of elevation of T from B is 32°.

Diagram not to scale

(a)

Calculate the distances FA and FB.

(3)

(b)

Calculate the distance AB.

(2)

(c)

M is the midpoint of AB. Calculate the angle of elevation of T from M.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: FA = 40 / tan 25° or FB = 40 / tan 32°	✓ 1
A1: FA = 85.8 m (85.78…)	✓ 1
A1: FB = 64.0 m (64.01…)	✓ 1
Part (b)
M1: AB² = 85.78² + 64.01², using angle AFB = 90°	✓ 1
A1: AB = 107 m (107.0…)	✓ 1
Part (c)
B1: FM = ½ × AB = 53.5 m, because the midpoint of the hypotenuse is equidistant from all three vertices	✓ 1
M1: tan θ = 40 / their 53.5	✓ 1
A1: θ = 36.8°	✓ 1

Exam tip: In 3D problems, sketch each right-angled triangle flat on its own. The ground triangle AFB is right-angled at F because south and east are perpendicular.

+ Add to test✓ In your test344IGMC-046-02-14Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EHard8 marks

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In triangle ABC, AB = x cm, AC = (x + 2) cm, BC = 7 cm and angle BAC = 60°.
You are given that cos 60° = ½ and sin 60° = \(\frac{\sqrt{3}}{2}\).

Diagram not to scale

(a)

Show that x² + 2x − 45 = 0.

(3)

(b)

Solve x² + 2x − 45 = 0 to find the exact value of x.

(2)

(c)

Show that the area of triangle ABC is \(\frac{45\sqrt{3}}{4}\) cm².

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: 7² = x² + (x + 2)² − 2x(x + 2) × ½	✓ 1
M1: 49 = x² + x² + 4x + 4 − x² − 2x	✓ 1
A1: x² + 2x − 45 = 0 with no errors seen	✓ 1
Part (b)
M1: (x + 1)² = 46, or the correct quadratic formula substitution	✓ 1
A1: x = −1 + √46, with the negative root rejected	✓ 1
Part (c)
M1: area = ½ × x(x + 2) × \(\frac{\sqrt{3}}{2}\)	✓ 1
M1: x(x + 2) = x² + 2x = 45 (from the equation in part (a))	✓ 1
A1: area = ½ × 45 × \(\frac{\sqrt{3}}{2}\) = \(\frac{45\sqrt{3}}{4}\)	✓ 1

Exam tip: Look for links between parts. In part (c), x(x + 2) equals x² + 2x, which part (a) shows is 45. This saves expanding the surd answer from part (b).

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Two rescue boats leave the same harbour H at the same time. Boat A travels 12 km on a bearing of 035°. Boat B travels 9 km on a bearing of 140°. Use unrounded values in your working.

Diagram not to scale

(a)

Show that angle AHB = 105°.

(1)

(b)

Calculate the distance AB. Give your answer correct to 3 significant figures.

(3)

(c)

Calculate the bearing of B from A. Give your answer correct to 1 decimal place.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 140° − 35° = 105°	✓ 1
Part (b)
M1 AB² = 12² + 9² − 2 × 12 × 9 × cos 105°	✓ 1
M1 AB² = 280.9	✓ 1
A1 16.8 km	✓ 1
Part (c)
M1 sin HAB ÷ 9 = sin 105° ÷ AB (ECF from (b))	✓ 1
A1 angle HAB = 31.2°	✓ 1
B1 bearing of H from A is 035° + 180° = 215°	✓ 1
A1 215° − 31.2° = 183.8° (accept 183.7 to 183.8)	✓ 1

Common error: Angle HAB is not a bearing. Use the back bearing 215° of the harbour from A, then subtract the angle to reach the direction of B.

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A water tank is a cuboid 2.4 m long, 1.8 m wide and 1.5 m high. Opposite corners are joined by straight rods through the inside of the tank.

Diagram not to scale

(a)

Calculate the length of a diagonal of the base.

(2)

(b)

Calculate the length of a rod joining opposite corners of the tank, to 3 significant figures.

(2)

(c)

Calculate the angle between a rod and the base of the tank, to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(2.4² + 1.8²)	✓ 1
A1 3.0 m	✓ 1
Part (b)
M1 √(3.0² + 1.5²) (ECF from (a))	✓ 1
A1 3.35 m	✓ 1
Part (c)
M1 tan⁻¹(1.5/3.0)	✓ 1
A1 26.6°	✓ 1

Exam tip: In a 3D problem, draw the right-angled triangle you are using as a flat sketch with its sides labelled.

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Three pumping stations A, B and C are in a water network. B is 7.5 km from A on a bearing of 050°. C is 9 km from B on a bearing of 160°. Use unrounded values in your working.

Diagram not to scale

(a)

Show that angle ABC = 70°.

(2)

(b)

Calculate AC, to 3 significant figures.

(3)

(c)

Calculate angle BAC, to 1 decimal place.

(3)

(d)

Find the bearing of C from A, to 1 decimal place.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of A from B is 050° + 180° = 230°	✓ 1
A1 230° − 160° = 70°	✓ 1
Part (b)
M1 AC² = 7.5² + 9² − 2 × 7.5 × 9 × cos 70°	✓ 1
A1 AC² = 91.1…	✓ 1
A1 9.54 km	✓ 1
Part (c)
M1 sin(BAC)/9 = sin 70°/AC	✓ 1
M1 BAC = sin⁻¹(9 sin 70° ÷ AC) (BAC is acute as it is not the largest angle)	✓ 1
A1 62.4°	✓ 1
Part (d)
B1 050° + 62.4° = 112.4° (ECF from (c))	✓ 1

Common error: Sketch the north line at each point before adding or subtracting bearings; the angle at a vertex is rarely the difference of two bearings without a back-bearing.

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A triangular zone in a game map has sides of 7 m, 9 m and 12 m. Use unrounded values in your working.

Diagram not to scale

(a)

Calculate the largest angle of the triangle, to 1 decimal place.

(3)

(b)

Calculate the area of the zone, to 3 significant figures.

(2)

(c)

Calculate the shortest distance from the vertex opposite the 12 m side to that side, to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cos θ = (7² + 9² − 12²)/(2 × 7 × 9)	✓ 1
A1 cos θ = −14/126 = −0.111…	✓ 1
A1 96.4°	✓ 1
Part (b)
M1 ½ × 7 × 9 × sin 96.4° (unrounded angle accepted)	✓ 1
A1 31.3 m²	✓ 1
Part (c)
M1 ½ × 12 × h = area (ECF from (b))	✓ 1
A1 5.22 m	✓ 1

Exam tip: The largest angle is opposite the longest side, so the cosine rule gives a clear starting point.

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A grain tower stands on level ground. From a point P on the ground, the angle of elevation of the top of the tower is 32°. From a point Q, which is on the same horizontal line as P and the base of the tower and is 30 m closer to the tower, the angle of elevation of the top is 44°. Let d m be the distance from P to the base of the tower and h m the height of the tower. Ignore the height of the observer.

Diagram not to scale

(a)

Show that d = (30 tan 44°)/(tan 44° − tan 32°).

(3)

(b)

Calculate the height h of the tower. Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
h = d tan 32° and h = (d − 30) tan 44°	✓ 1
Equates: d tan 32° = d tan 44° − 30 tan 44°	✓ 1
Rearranges to d(tan 44° − tan 32°) = 30 tan 44°, giving the result	✓ 1
Part (b)
d = 85.0 m (unrounded value used)	✓ 1
h = d tan 32° = 53.1 m	✓ 1

Exam tip: When two right-angled triangles share a side, write one equation for that shared side from each triangle and equate them.

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An air-traffic radar at O shows aircraft P on a bearing of 040° at a distance of 120 km and aircraft Q on a bearing of 125° at a distance of 85 km. Both aircraft are at the same altitude.

Diagram not to scale

(a)

Calculate the distance PQ. Give your answer to 3 significant figures.

(3)

(b)

Calculate the bearing of Q from P. Angle OPQ is acute. Give your answer to 1 decimal place.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Angle POQ = 125° − 40° = 85°	✓ 1
Cosine rule: PQ² = 120² + 85² − 2 × 120 × 85 × cos 85°	✓ 1
PQ = 141 km	✓ 1
Part (b)
Sine rule: sin OPQ/85 = sin 85°/PQ, ECF from (a)	✓ 1
Angle OPQ = 36.9°	✓ 1
Bearing of O from P is 040° + 180° = 220°	✓ 1
Bearing of Q from P = 220° − 36.9° = 183.1°	✓ 1

Common error: Subtract the bearings to get the angle at the radar, but remember that a bearing from the other end of a line is found by adding or subtracting 180°.

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A ship leaves port P and sails 40 km on a bearing of 050° to point A. It then sails 30 km on a bearing of 110° to point B.

Diagram not to scale

(a)

Show that angle PAB = 120°.

(2)

(b)

Calculate the distance PB. Give your answer to 3 significant figures.

(3)

(c)

Calculate the bearing of B from P. Angle APB is acute. Give your answer to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Bearing of P from A is 50° + 180° = 230°	✓ 1
230° − 110° = 120°	✓ 1
Part (b)
Cosine rule: PB² = 40² + 30² − 2 × 40 × 30 × cos 120°	✓ 1
PB² = 3700	✓ 1
PB = 60.8 km	✓ 1
Part (c)
Sine rule: sin APB/30 = sin 120°/PB, ECF from (b)	✓ 1
Angle APB = 25.3°	✓ 1
Bearing = 50° + 25.3° = 75.3°	✓ 1

Exam tip: When the included angle is obtuse, cos is negative, so the cosine rule adds the product term; check that your side is longer than both known sides.

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In a harbour the depth of water d metres is modelled by d = 6 + 2.5 cos(30t)°, where t is the time in hours after high tide and the angle is in degrees.

(a)

Write down the maximum depth and the minimum depth of the water.

(2)

(b)

Write down the time between one high tide and the next.

(1)

(c)

Find the first time after high tide when the depth is exactly 5 m. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Maximum depth = 6 + 2.5 = 8.5 m	✓ 1
Minimum depth = 6 − 2.5 = 3.5 m	✓ 1
Part (b)
Period = 360/30 = 12 hours	✓ 1
Part (c)
Sets 6 + 2.5 cos(30t) = 5 so cos(30t) = −0.4	✓ 1
30t = 113.6° (first solution after 0)	✓ 1
t = 3.79 hours	✓ 1

Exam tip: For the first positive solution of cos x = negative number, the calculator's principal value is the one you need; it lies between 90° and 180°.

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A data hall is a cuboid. The floor ABCD is a rectangle with AB = 15 m and BC = 9 m. The hall is 4 m high, and EFGH is the ceiling, with E directly above A, F above B, G above C and H above D. A cable runs in a straight line from A to G.

Diagram not to scale

(a)

Calculate the length of AC. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the length of the cable AG. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the angle between the cable AG and the floor ABCD. Give your answer correct to 1 decimal place.

(2)

(d)

Calculate the angle between AG and the edge AD. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 use of Pythagoras in triangle ABC, 15² + 9²	✓ 1
A1 17.5 m	✓ 1
Part (b)
M1 use of Pythagoras in triangle ACG, using the unrounded AC or AC² = 306 and CG = 4	✓ 1
A1 17.9 m	✓ 1
Part (c)
M1 tan (angle GAC) = 4 ÷ AC	✓ 1
A1 12.9°	✓ 1
Part (d)
M1 triangle ADG has a right angle at D, so cos (angle DAG) = 9 ÷ AG (ECF from (b))	✓ 1
A1 59.9°	✓ 1

Exam tip: In 3D problems, draw the right-angled triangle that contains the angle you need, and name its three vertices before you calculate. Work with unrounded lengths when you can.

+ Add to test✓ In your test354IGMC3-6-09r7-12Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EHard9 marks

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Two search teams leave a base camp C at the same time. Team A walks 9.0 km on a bearing of 038°. Team B walks 14.0 km on a bearing of 112°. The ground is level.

Diagram not to scale

(a)

Show that angle ACB = 74°.

(1)

(b)

Calculate the distance AB between the teams. Give your answer correct to 3 significant figures.

(3)

(c)

Calculate angle CAB. Give your answer correct to 1 decimal place.

(3)

(d)

Calculate the bearing of team B from team A. Give your answer correct to the nearest degree.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 112° − 038° = 74°	✓ 1
Part (b)
M1 cosine rule, AB² = 9² + 14² − 2 × 9 × 14 × cos 74°	✓ 1
M1 AB² = 207.5…	✓ 1
A1 14.4 km	✓ 1
Part (c)
M1 sine rule, sin CAB ÷ 14 = sin 74° ÷ AB (ECF from (b))	✓ 1
M1 sin CAB = 0.934…	✓ 1
A1 69.1° (the obtuse value is rejected because it would make the angle sum exceed 180°)	✓ 1
Part (d)
M1 bearing of C from A is 038° + 180° = 218°, and B lies anticlockwise from this direction, so subtract angle CAB	✓ 1
A1 218° − 69.1° = 149° (ECF from (c))	✓ 1

Exam tip: When the sine rule gives an angle, check whether the obtuse alternative is possible by testing the angle sum of the triangle. Here only the acute value fits.

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The glass roof of a library atrium is a square-based pyramid. The base ABCD is a horizontal square of side 8.0 m. The apex V is 3.0 m vertically above the centre of the base.

Diagram not to scale

(a)

Calculate the length of a sloping edge, such as VA. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate the angle between VA and the base. Give your answer correct to 1 decimal place.

(2)

(c)

Calculate the height of one triangular glass face, measured from V to the midpoint of a base edge.

(2)

(d)

The glass costs €85 per square metre. Calculate the cost of the glass for the four faces.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 half the diagonal of the base is √(4² + 4²) = √32 (5.66…)	✓ 1
M1 VA² = 32 + 3² = 41	✓ 1
A1 6.40 m	✓ 1
Part (b)
M1 tan x = 3 ÷ √32	✓ 1
A1 27.9°	✓ 1
Part (c)
M1 3² + 4² (the horizontal distance from the centre to the midpoint of an edge is 4 m)	✓ 1
A1 5 m	✓ 1
Part (d)
M1 area of four faces = 4 × ½ × 8 × 5 = 80 m²	✓ 1
A1 €6800 (ECF from (c))	✓ 1

Exam tip: In a pyramid, use two different right-angled triangles: one through the centre of the base for edge lengths and one through the midpoint of a base edge for face heights.

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From a lighthouse L, turbine A is 6.5 km away on a bearing of 148° and turbine B is 8.2 km away on a bearing of 232°. The sea is calm and the lighthouse and turbines are at sea level.

Diagram not to scale

(a)

Show that angle ALB = 84°.

(1)

(b)

Calculate the distance AB. Give your answer correct to 3 significant figures.

(3)

(c)

Calculate angle LBA. Give your answer correct to 1 decimal place.

(2)

(d)

Calculate the bearing of A from B. Give your answer correct to the nearest degree.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 232° − 148° = 84°	✓ 1
Part (b)
M1 cosine rule, AB² = 6.5² + 8.2² − 2 × 6.5 × 8.2 × cos 84°	✓ 1
M1 AB² = 99.9…	✓ 1
A1 9.92 km	✓ 1
Part (c)
M1 sine rule, sin LBA ÷ 6.5 = sin 84° ÷ AB (ECF from (b))	✓ 1
A1 40.7°	✓ 1
Part (d)
M1 bearing of L from B is 232° − 180° = 052°, and A lies clockwise from this direction, so add angle LBA	✓ 1
A1 052° + 40.7° = 093° (ECF from (c))	✓ 1

Common error: The sine rule gives an angle but not a bearing. Add or subtract the angle from a known bearing, using a sketch to decide which.

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A summit is modelled as a pyramid. The base ABCD is a horizontal rectangle with AB = 120 m and BC = 90 m. The apex V is vertically above the centre M of the base, with VM = 150 m.

Diagram not to scale

(a)

Calculate the length of the diagonal AC.

(2)

(b)

Calculate the length of the edge VA. Give your answer to 3 significant figures.

(2)

(c)

Calculate the angle between VA and the base. Give your answer to 1 decimal place.

(2)

(d)

Calculate angle AVC. Give your answer to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(120² + 90²)	✓ 1
A1 150 m	✓ 1
Part (b)
M1 √(75² + 150²)	✓ 1
A1 168 m	✓ 1
Part (c)
M1 tan⁻¹(150/75)	✓ 1
A1 63.4°	✓ 1
Part (d)
M1 cosine rule in triangle AVC with VA = VC	✓ 1
M1 cos AVC = (2 × VA² − 150²) ÷ (2 × VA²) = 0.6	✓ 1
A1 53.1°	✓ 1

Exam tip: In 3D problems, draw the right-angled or isosceles triangle that contains the required angle and label its sides first.

+ Add to test✓ In your test358IGMC3-6-10r8-14Pythagoras' theorem and right-angled trigonometry Paper EHard7 marks

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A triangular pastry cutter ABC has AB = 9.5 cm, AC = 7.0 cm and angle ABC = 44°.

(a)

Show that sin ACB = 0.943 correct to 3 significant figures.

(2)

(b)

Find the two possible sizes of angle ACB. Give your answers to 1 decimal place.

(2)

(c)

The cutter has an obtuse angle at C. Calculate the length of BC, using unrounded values. Give your answer to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sin C/9.5 = sin 44°/7.0	✓ 1
A1 sin C = 9.5 × sin 44° ÷ 7.0 = 0.943	✓ 1
Part (b)
B1 70.5°	✓ 1
B1 180° − 70.5° = 109.5°	✓ 1
Part (c)
M1 angle BAC = 180° − 44° − 109.5° = 26.5°	✓ 1
M1 BC/sin A = 7.0/sin 44°	✓ 1
A1 4.50 cm	✓ 1

Common error: Giving only the acute value from sin⁻¹; check whether 180° minus that angle also fits the triangle.

+ Add to test✓ In your test359IGMC3-6-09r9-12Pythagoras' theorem and right-angled trigonometry Paper EHard9 marks

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A horizontal rectangular platform ABCD has AB = 24 m and BC = 18 m. A vertical mast AT of height 13 m stands at corner A, with T at the top. A cable is stretched from T to corner C. Use unrounded values in your working; give lengths to 3 significant figures and angles to 1 decimal place.

Diagram not to scale

(a)

Calculate AC.

(2)

(b)

Calculate the length of the cable TC.

(2)

(c)

Calculate the angle between the cable TC and the platform.

(2)

(d)

Calculate angle BTC.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 AC² = 24² + 18²	✓ 1
A1 30 m	✓ 1
Part (b)
M1 TC² = 30² + 13² (= 1069)	✓ 1
A1 32.7 m	✓ 1
Part (c)
M1 angle TCA, tan = 13 ÷ 30	✓ 1
A1 23.4°	✓ 1
Part (d)
M1 TB = √(24² + 13²) = √745	✓ 1
M1 BC is perpendicular to TB so angle TBC = 90°, tan(BTC) = 18 ÷ √745	✓ 1
A1 33.4°	✓ 1

Exam tip: In 3D problems, find the right-angled triangle that contains the angle, then name its vertices carefully.

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A coffee freighter leaves port P and sails 40 km on a bearing of 070° to buoy A. A second vessel leaves P and sails 55 km on a bearing of 130° to buoy B.

Diagram not to scale

(a)

Show that angle APB = 60°.

(1)

(b)

Calculate the distance AB. Give your answer correct to 4 significant figures.

(3)

(c)

Use AB = 49.24 km to calculate angle PAB. Give your answer correct to 1 decimal place.

(3)

(d)

Find the bearing of B from A. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 130° − 70° = 60°	✓ 1
Part (b)
M1 AB² = 40² + 55² − 2 × 40 × 55 × cos 60°	✓ 1
A1 AB² = 2425	✓ 1
A1 49.24 km	✓ 1
Part (c)
M1 cosine rule: cos A = (40² + 49.24² − 55²) ÷ (2 × 40 × 49.24)	✓ 1
A1 cos A = 0.2538	✓ 1
A1 75.3°	✓ 1
Part (d)
M1 bearing of P from A = 250°	✓ 1
M1 250° − 75.3° (ECF from (c))	✓ 1
A1 174.7°	✓ 1

Common error: Subtracting angles from the wrong back bearing. Draw a north line at A and mark the bearing of P from A first.

+ Add to test✓ In your test361IGMC3-6-09r10-12Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EHard12 marks

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A tent for a music festival is a square-based pyramid VABCD. The horizontal base ABCD has side 10 m. The vertex V is 12 m vertically above the centre of the base.

Diagram not to scale

(a)

Calculate the length of the diagonal AC. Give your answer to 3 significant figures.

(2)

(b)

Calculate the length of the sloping edge VA. Give your answer to 3 significant figures.

(3)

(c)

Calculate the angle between VA and the base. Give your answer to 1 decimal place.

(2)

(d)

M is the midpoint of AB and N is the centre of the base. Show that VM = 13 m, and hence calculate the cost of fabric for the four triangular faces at €3.20 per square metre.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 AC = √(10² + 10²)	✓ 1
A1 14.1 m	✓ 1
Part (b)
B1 half diagonal = 7.07… m	✓ 1
M1 VA² = 12² + 7.07…² (= 194)	✓ 1
A1 13.9 m	✓ 1
Part (c)
M1 tan x = 12/7.07…	✓ 1
A1 59.5°	✓ 1
Part (d)
B1 NM = 5 m	✓ 1
M1 VM² = 12² + 5² = 169, so VM = 13 m	✓ 1
M1 total area = 4 × ½ × 10 × 13 = 260 m²	✓ 1
M1 cost = 260 × 3.20	✓ 1
A1 €832	✓ 1

Exam tip: In 3D problems, identify a right-angled triangle that contains the angle you want and redraw it flat.

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The displacement y mm of a point on a vibrating string is modelled by y = 3 sin(4x°), where x is a scaled time value with 0 ≤ x ≤ 90.

(a)

Write down the amplitude of the motion.

(1)

(b)

Write down the period of the function, in terms of x.

(1)

(c)

Solve 3 sin(4x°) = 1.5 for 0 ≤ x ≤ 90.

(3)

(d)

Solve 3 sin(4x°) = −2.4 for 0 ≤ x ≤ 90. Give your answers to 1 decimal place.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 3 mm	✓ 1
Part (b)
B1 90	✓ 1
Part (c)
M1 sin(4x°) = 0.5	✓ 1
A1 4x = 30 or 150, so x = 7.5	✓ 1
A1 x = 37.5	✓ 1
Part (d)
M1 sin(4x°) = −0.8	✓ 1
M1 4x = 180 + 53.13… = 233.13… (or 360 − 53.13… = 306.87…)	✓ 1
A1 x = 58.3	✓ 1
A1 x = 76.7	✓ 1

Common error: After finding the first angle, divide every solution by the coefficient of x; check that you have found all solutions in the interval 0 ≤ 4x ≤ 360.

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Two monitoring stations P and Q lie on level ground. Q is 2.4 km due east of P. The summit V of a volcano is seen from P on a bearing of 040° and from Q on a bearing of 325°. Use unrounded values in each calculation and give distances to 3 significant figures unless told otherwise.

Diagram not to scale

(a)

Show that angle VPQ = 50° and angle VQP = 55°.

(2)

(b)

Calculate the horizontal distance PV.

(3)

(c)

From P the angle of elevation of the summit is 21°. Calculate the height of the summit above the level of the stations, in metres, to the nearest metre.

(2)

(d)

Calculate the angle of elevation of the summit from Q. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 angle VPQ = 090° − 040° = 50°	✓ 1
A1 bearing of P from Q is 270°, so angle VQP = 325° − 270° = 55°	✓ 1
Part (b)
B1 angle PVQ = 180° − 50° − 55° = 75°	✓ 1
M1 PV ÷ sin 55° = 2.4 ÷ sin 75°	✓ 1
A1 2.04 km	✓ 1
Part (c)
M1 height = PV × tan 21° (PV = 2.0353 km)	✓ 1
A1 781 m	✓ 1
Part (d)
M1 QV = 2.4 × sin 50° ÷ sin 75° = 1.903 km and tan⁻¹(0.7813 ÷ 1.903) (ECF from (c))	✓ 1
A1 22.3°	✓ 1

Exam tip: Convert bearings into interior angles with a sketch of the north lines before using the sine rule.

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A model of a volcano is a solid pyramid with a horizontal square base of side 60 cm. The vertex V is 40 cm vertically above the centre of the base.

Diagram not to scale

(a)

Calculate the length of a sloping edge of the pyramid. Give your answer to 3 significant figures.

(3)

(b)

Calculate the angle between a sloping edge and the base. Give your answer correct to 1 decimal place.

(2)

(c)

Calculate the angle between a sloping triangular face and the base. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 distance from centre of base to a corner = ½ × sqrt(60² + 60²) = 42.4	✓ 1
M1 sqrt(42.4² + 40²) (or sqrt(1800 + 1600))	✓ 1
A1 58.3 cm	✓ 1
Part (b)
M1 tan⁻¹(40 ÷ 42.4) (ECF from (a))	✓ 1
A1 43.3°	✓ 1
Part (c)
B1 horizontal distance from the centre to the midpoint of a base edge = 30 cm	✓ 1
M1 tan⁻¹(40 ÷ 30)	✓ 1
A1 53.1°	✓ 1

Common error: The angle between a face and the base uses the midpoint of a base edge, not a corner of the base.

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A ship leaves port A and sails 36 km on a bearing of 047° to a point B. It then sails 28 km on a bearing of 158° to a point C. Use unrounded values in each calculation.

Diagram not to scale

(a)

Show that angle ABC = 69°.

(2)

(b)

Calculate the distance AC. Give your answer to 3 significant figures.

(3)

(c)

Calculate the bearing of C from A. Give your answer as a three-figure bearing correct to 1 decimal place.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of A from B = 047° + 180° = 227°	✓ 1
A1 227° − 158° = 69°	✓ 1
Part (b)
M1 AC² = 36² + 28² − 2 × 36 × 28 × cos 69°	✓ 1
A1 AC² = 1358 (1357.5...)	✓ 1
A1 AC = 36.8 km	✓ 1
Part (c)
M1 sin(BAC) ÷ 28 = sin 69° ÷ AC	✓ 1
A1 angle BAC = 45.2°	✓ 1
M1 bearing = 047° + 45.2°	✓ 1
A1 092.2°	✓ 1

Exam tip: Check the turn direction: here the ship turns right, so C lies clockwise of B as seen from A.

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In a game, a player walks from base camp B on a bearing of 038° for 640 m to a flag F. A second player walks from B on a bearing of 112° for 410 m to a chest C. Give lengths to 3 significant figures and angles to 1 decimal place. Use unrounded values in later parts.

Diagram not to scale

(a)

Show that angle FBC = 74°.

(1)

(b)

Calculate the distance FC.

(3)

(c)

Calculate the bearing of C from F.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 112° − 38° = 74°	✓ 1
Part (b)
M1 cosine rule FC² = 640² + 410² − 2 × 640 × 410 × cos 74°	✓ 1
M1 FC² = 433 000 (approx.) or correct evaluation	✓ 1
A1 658 m	✓ 1
Part (c)
M1 sine rule sin(BFC)/410 = sin 74° / FC	✓ 1
A1 angle BFC = 36.8°	✓ 1
M1 bearing of B from F = 038° + 180° = 218°, then 218° − angle BFC	✓ 1
A1 181.2°	✓ 1

Exam tip: Use the sine rule for an angle only when the angle you want is acute or you have checked it cannot be obtuse; here BFC is opposite the shorter side.

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Calculator

A pyramid in a game has a square base ABCD of side 80 m. The vertex V is vertically above the centre O of the base, and VO = 55 m. Give lengths to 3 significant figures and angles to 1 decimal place. Use unrounded values in later parts.

Diagram not to scale

(a)

Calculate the length of the diagonal AC.

(2)

(b)

Calculate the length of the sloping edge VA.

(3)

(c)

Calculate the angle between VA and the base.

(2)

(d)

M is the midpoint of AB. Calculate the angle between the face VAB and the base.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sqrt(80² + 80²)	✓ 1
A1 113 m	✓ 1
Part (b)
M1 AO = half of AC = 56.6… m	✓ 1
M1 VA² = AO² + 55²	✓ 1
A1 78.9 m	✓ 1
Part (c)
M1 tan(VAO) = 55 / AO	✓ 1
A1 44.2°	✓ 1
Part (d)
B1 OM = 40 m	✓ 1
M1 tan(VMO) = 55/40	✓ 1
A1 54.0°	✓ 1

Common error: The angle between a face and the base is measured along the line of greatest slope (through the midpoint of the base edge), not along an edge.

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After a lorry crosses a footbridge, the vertical displacement h millimetres of the middle of the deck from its rest position is modelled by h = 3.2 sin(72t)°, where t is the time in seconds after the lorry leaves the bridge.

(a)

Write down the amplitude of the oscillation.

(1)

(b)

Find the period of the oscillation.

(2)

(c)

Calculate h when t = 1.5. Give your answer correct to 3 significant figures.

(2)

(d)

Find the first time when h = 2.0. Give your answer correct to 3 significant figures.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 3.2 mm	✓ 1
Part (b)
M1 360° / 72	✓ 1
A1 5 s	✓ 1
Part (c)
M1 h = 3.2 sin(72 × 1.5) = 3.2 sin 108°	✓ 1
A1 3.04 mm	✓ 1
Part (d)
M1 sin(72t) = 2.0 / 3.2 = 0.625	✓ 1
M1 72t = sin⁻¹(0.625) = 38.68°	✓ 1
A1 t = 0.537 s	✓ 1

Common error: With a graph model such as sin(72t), the angle 72t is in degrees, so the calculator must be in degree mode.

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A coffee storage crate is a cuboid with a horizontal rectangular base 60 cm long and 45 cm wide. The crate is 36 cm high. A straight rod is placed from one bottom corner to the opposite top corner of the crate.

Diagram not to scale

(a)

Calculate the length of a diagonal of the base.

(2)

(b)

Calculate the length of the rod. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the angle between the rod and the base of the crate. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sqrt(60² + 45²)	✓ 1
A1 75 cm	✓ 1
Part (b)
M1 sqrt(75² + 36²)	✓ 1
A1 83.2 cm	✓ 1
Part (c)
M1 tan x = 36 / 75	✓ 1
A1 25.6°	✓ 1

Exam tip: For an angle between a line and a plane, use the right-angled triangle formed by the line, its projection on the plane and the vertical.

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A field hospital tent is a pyramid VABCD. The base ABCD is a horizontal rectangle with AB = 12 m and BC = 9 m. The apex V is vertically above the centre O of the base. Each sloping edge VA, VB, VC and VD has length 8.5 m.

Diagram not to scale

(a)

Calculate the length of the diagonal AC.

(2)

(b)

Calculate the height VO of the tent.

(3)

(c)

Calculate the angle between the edge VA and the base ABCD. Give your answer correct to 1 decimal place.

(2)

(d)

M is the midpoint of AB. Calculate the angle between the triangular face VAB and the base ABCD. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{12^2+9^2}\)	✓ 1
A1 15 m	✓ 1
Part (b)
B1 \(AO=7.5\) m	✓ 1
M1 \(VO=\sqrt{8.5^2-7.5^2}\)	✓ 1
A1 4 m	✓ 1
Part (c)
M1 \(\cos^{-1}\left(\frac{7.5}{8.5}\right)\) or \(\tan^{-1}\left(\frac{4}{7.5}\right)\) (angle VAO)	✓ 1
A1 28.1°	✓ 1
Part (d)
B1 angle VMO identified, with OM = 4.5 m (half of BC)	✓ 1
M1 \(\tan^{-1}\left(\frac{4}{4.5}\right)\)	✓ 1
A1 41.6°	✓ 1

Exam tip: In a 3D problem, draw the right-angled triangle you need (here VOA or VOM) as a separate flat sketch and label its sides.

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The number of new infections per day in a town is modelled by \(N=80+35\sin(15t)^\circ\), where \(t\) is the number of weeks after 1 January and \(0\le t\le24\). Give all non-exact answers correct to 3 significant figures.

(a)

Write down the maximum and minimum values of \(N\).

(2)

(b)

Find the period of the model in weeks.

(1)

(c)

Solve \(N=100\) for \(0\le t\le24\).

(4)

(d)

Solve \(N=60\) for \(0\le t\le24\).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 maximum 115	✓ 1
B1 minimum 45	✓ 1
Part (b)
B1 \(\frac{360}{15}=24\) weeks	✓ 1
Part (c)
M1 \(\sin(15t)=\frac{20}{35}\)	✓ 1
A1 \(15t=34.8^\circ\) (awrt)	✓ 1
M1 second solution using \(180^\circ-34.8^\circ\)	✓ 1
A1 \(t=2.32\) and \(t=9.68\)	✓ 1
Part (d)
M1 \(\sin(15t)=-\frac{20}{35}\)	✓ 1
M1 \(15t=180^\circ+34.8^\circ\) and \(15t=360^\circ-34.8^\circ\)	✓ 1
A1 \(t=14.3\) and \(t=21.7\)	✓ 1

Common error: After finding the first angle from the inverse sine, forgetting the second solution in the range; for a negative sine value use \(180^\circ+\alpha\) and \(360^\circ-\alpha\).

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Calculator

A shed roof is a pyramid VABCD. The base ABCD is a horizontal square of side 12 m. The vertex V is 8 m vertically above the centre M of the base. Give answers to 3 significant figures where they are not exact, and use unrounded values in later parts.

Diagram not to scale

(a)

Calculate the length of the diagonal AC.

(2)

(b)

Calculate the length of the sloping edge VA.

(2)

(c)

Calculate the angle between VA and the base ABCD.

(2)

(d)

N is the midpoint of AB. Calculate the angle between the face VAB and the base ABCD.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 12² + 12² = 288	✓ 1
A1 17.0 m (16.97… m)	✓ 1
Part (b)
M1 VA² = 8² + (½ AC)² = 64 + 72	✓ 1
A1 11.7 m	✓ 1
Part (c)
M1 tan x = 8 ÷ 8.485… (or 8 ÷ (6√2))	✓ 1
A1 43.3°	✓ 1
Part (d)
M1 angle VNM with tan y = 8 ÷ 6	✓ 1
A1 53.1°	✓ 1

Exam tip: In a 3D problem, draw the right-angled triangle you need as a flat picture and label which lengths are horizontal and which are vertical.

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Calculator

A medical supplies crate is a cuboid ABCDEFGH. The base ABCD is a horizontal rectangle with AB = 60 cm and BC = 45 cm. The vertical edges AE, BF, CG and DH each have length 30 cm. Use unrounded values in later parts.

Diagram not to scale

(a)

Calculate the length of the diagonal AG. Give your answer correct to 3 significant figures.

(3)

(b)

Calculate the angle between AG and the base ABCD. Give your answer correct to 1 decimal place.

(3)

(c)

Calculate the angle between the plane ABGH and the base ABCD.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 AC² = 60² + 45², so AC = 75 cm	✓ 1
M1 AG² = 75² + 30² (or 60² + 45² + 30²)	✓ 1
A1 80.8 cm	✓ 1
Part (b)
M1 angle GAC	✓ 1
M1 tan x = 30 ÷ 75	✓ 1
A1 21.8°	✓ 1
Part (c)
M1 angle GBC, tan y = 30 ÷ 45	✓ 1
A1 33.7°	✓ 1

Exam tip: The angle between a line and a plane is found using the line's projection onto the plane, so identify that projection first.

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Calculator

Two helicopters leave a hospital pad O at the same time. Helicopter H flies 25 km on a bearing of 040° and helicopter K flies 32 km on a bearing of 115°. Use unrounded values in later parts.

Diagram not to scale

(a)

Show that angle HOK = 75°.

(1)

(b)

Calculate the distance HK. Give your answer correct to 3 significant figures.

(3)

(c)

Calculate angle OHK, which is acute. Give your answer correct to 1 decimal place.

(3)

(d)

Calculate the bearing of K from H. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 115° − 40° = 75°	✓ 1
Part (b)
M1 HK² = 25² + 32² − 2 × 25 × 32 × cos 75°	✓ 1
M1 HK² = 1234.9…	✓ 1
A1 35.1 km	✓ 1
Part (c)
M1 sine rule: sin(OHK) ÷ 32 = sin 75° ÷ HK	✓ 1
M1 sin(OHK) = 32 sin 75° ÷ HK	✓ 1
A1 61.6°	✓ 1
Part (d)
M1 bearing of O from H is 220° (40° + 180°)	✓ 1
A1 220° − 61.6° = 158.4° (ECF from (c))	✓ 1

Exam tip: Find the bearing of O from H first; the required bearing is then that value plus or minus the angle inside the triangle, so sketch which way K lies.

+ Add to test✓ In your test375IGMC3-6-09r21-12Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EHard9 marks

Calculator

A model roof is a pyramid VABCD. The base ABCD is a horizontal square of side 12 cm. The apex V is vertically above the centre of the base at a height of 8 cm.

Diagram not to scale

(a)

Calculate the length of the diagonal AC, to 3 significant figures.

(2)

(b)

Calculate the length of the sloping edge VA, to 3 significant figures.

(2)

(c)

Calculate the angle between VA and the base, to 1 decimal place.

(2)

(d)

Calculate angle AVC, to 1 decimal place. Use \(VA^2=136\) and \(AC^2=288\).

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{12^2+12^2}\)	✓ 1
A1 17.0 cm	✓ 1
Part (b)
M1 \(\sqrt{8^2+\left(\frac{AC}{2}\right)^2}=\sqrt{64+72}\)	✓ 1
A1 11.7 cm	✓ 1
Part (c)
M1 \(\tan^{-1}\left(\frac{8}{\sqrt{72}}\right)\)	✓ 1
A1 43.3°	✓ 1
Part (d)
M1 \(\cos AVC=\frac{136+136-288}{2\times136}\)	✓ 1
A1 \(\cos AVC=-\frac{16}{272}\)	✓ 1
A1 93.4°	✓ 1

Exam tip: If the cosine comes out negative, the angle is obtuse; your calculator gives the correct obtuse angle directly.

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Calculator

A triangular field ABC on a farm has AB = 85 m, AC = 120 m and angle BAC = 68°.

Diagram not to scale

(a)

Calculate the length BC. Give your answer correct to 3 significant figures.

(4)

(b)

Calculate the area of the field. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the size of angle ACB. Use the unrounded value of BC and give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cosine rule BC² = 85² + 120² − 2 × 85 × 120 × cos 68°	✓ 1
A1 BC² = 13 983 (accept 13 980 or 13 983.0)	✓ 1
M1 take square root	✓ 1
A1 118 m	✓ 1
Part (b)
M1 ½ × 85 × 120 × sin 68°	✓ 1
A1 4730 m²	✓ 1
Part (c)
M1 sin C / 85 = sin 68° / BC	✓ 1
M1 C = sin⁻¹(85 × sin 68° ÷ 118.25)	✓ 1
A1 41.8°	✓ 1

Exam tip: Use the cosine rule when you know two sides and the angle between them.

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Calculator

An aircraft flies from airport A on a bearing of 040° for 220 km to airport B. It then flies on a bearing of 135° for 150 km to airport C. Use unrounded values in your working and round only your final answers.

Diagram not to scale

(a)

Show that angle ABC = 85°.

(2)

(b)

Calculate the distance AC. Give your answer correct to 3 significant figures.

(4)

(c)

Calculate the bearing of C from A. Give your answer correct to 1 decimal place.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 bearing of A from B is 220°	✓ 1
A1 220 − 135 = 85°	✓ 1
Part (b)
M1 AC² = 220² + 150² − 2 × 220 × 150 × cos 85°	✓ 1
A1 AC² = 65 148 (accept 65 150)	✓ 1
M1 square root	✓ 1
A1 255 km	✓ 1
Part (c)
M1 sine rule sin A / 150 = sin 85° / AC	✓ 1
A1 angle BAC = 35.8°	✓ 1
M1 bearing = 040° + angle BAC	✓ 1
A1 075.8°	✓ 1

Common error: Add the angle found to the 040° bearing: bearings are always measured clockwise from north.

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Calculator

The roof of a barn is a pyramid with a horizontal square base ABCD of side 8 m. The apex V is vertically above the centre of the base at a height of 3 m. Use unrounded values in your working and round only your final answers.

Diagram not to scale

(a)

Calculate the length of the diagonal AC. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the length of the sloping edge VA. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the angle between VA and the base. Give your answer correct to 1 decimal place.

(2)

(d)

Calculate angle AVB. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 sqrt(8² + 8²)	✓ 1
A1 11.3 m	✓ 1
Part (b)
M1 sqrt((AC/2)² + 3²) = sqrt(32 + 9)	✓ 1
A1 6.40 m	✓ 1
Part (c)
M1 tan x = 3 / (AC/2) = 3 / 5.657	✓ 1
A1 27.9°	✓ 1
Part (d)
M1 cosine rule cos V = (VA² + VB² − AB²) / (2 × VA × VB)	✓ 1
M1 (41 + 41 − 64) / 82	✓ 1
A1 77.3°	✓ 1

Exam tip: Triangle VAB is isosceles, so VA = VB; write this down before using the cosine rule.

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Two research ships, \(A\) and \(B\), are 12.5 km apart. The bearing of \(B\) from \(A\) is 072°. A whale at \(W\) is on a bearing of 115° from \(A\) and on a bearing of 190° from \(B\).

Diagram not to scale

(a)

Show that angle \(AWB = 75°\).

(3)

(b)

Calculate the distance \(BW\), to 3 significant figures.

(3)

(c)

Calculate the shortest distance from \(W\) to the straight line \(AB\). Use the unrounded value of \(BW\). Give your answer to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 angle \(BAW = 115° - 72° = 43°\)	✓ 1
B1 bearing of \(A\) from \(B\) is \(252°\), so angle \(ABW = 252° - 190° = 62°\)	✓ 1
B1 \(180° - 43° - 62° = 75°\)	✓ 1
Part (b)
M1 sine rule \(\frac{BW}{\sin 43°} = \frac{12.5}{\sin 75°}\)	✓ 1
M1 \(BW = \frac{12.5 \sin 43°}{\sin 75°}\)	✓ 1
A1 8.83 km	✓ 1
Part (c)
M1 \(BW \sin 62°\)	✓ 1
A1 7.79 km	✓ 1

Exam tip: Label each triangle angle on a sketch with its working (for example 252° − 190°) before applying the sine rule; each side must be paired with the opposite angle.

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The roof of a glass pavilion is a pyramid with a horizontal square base \(ABCD\) of side 10 m. The vertex \(V\) is vertically above the centre \(M\) of the base. Each sloping edge, such as \(VA\), is 13 m long.

Diagram not to scale

(a)

Calculate the length of the diagonal \(AC\). Give your answer to 3 significant figures.

(2)

(b)

Calculate the vertical height \(VM\) of the pavilion. Use the unrounded value of \(AM\). Give your answer to 3 significant figures.

(3)

(c)

Calculate the angle between the edge \(VA\) and the base, to 1 decimal place.

(2)

(d)

Calculate angle \(AVC\), to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{10^2 + 10^2}\)	✓ 1
A1 14.1 m	✓ 1
Part (b)
B1 \(AM = \frac{\sqrt{200}}{2} = 7.07\ldots\) m	✓ 1
M1 \(VM = \sqrt{13^2 - AM^2} = \sqrt{169 - 50}\)	✓ 1
A1 10.9 m	✓ 1
Part (c)
M1 \(\cos^{-1}\left(\frac{AM}{13}\right)\) (or \(\sin^{-1}\left(\frac{VM}{13}\right)\))	✓ 1
A1 57.0°	✓ 1
Part (d)
M1 cosine rule \(\cos AVC = \frac{13^2 + 13^2 - 200}{2 \times 13 \times 13}\)	✓ 1
M1 evaluates \(\frac{138}{338}\)	✓ 1
A1 65.9°	✓ 1

Exam tip: In pyramid problems, identify the right-angled triangle containing the vertical height: its other two sides are half the base diagonal and the sloping edge.

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Two straight cables, PA and PB, run from a point P on a bridge pylon to two points A and B on the bridge deck. PA = 110 m, PB = 85 m and angle APB = 56°.

Diagram not to scale

(a)

Calculate the length AB. Give your answer correct to 3 significant figures.

(3)

(b)

Use the sine rule to calculate angle PAB. Give your answer correct to 1 decimal place.

(3)

(c)

Explain why angle PAB cannot be obtuse.

(1)

(d)

Calculate the area of triangle APB. Give your answer correct to the nearest square metre.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 AB² = 110² + 85² − 2 × 110 × 85 × cos 56°	✓ 1
A1 AB² = 8868 (to 4 significant figures)	✓ 1
A1 AB = 94.2 m	✓ 1
Part (b)
M1 sin A ÷ 85 = sin 56° ÷ AB	✓ 1
M1 sin A = 85 × sin 56° ÷ AB	✓ 1
A1 48.4°	✓ 1
Part (c)
R1 PB is shorter than PA, so angle PAB is smaller than angle PBA; the two angles add up to 124°, so angle PAB is less than 62°	✓ 1
Part (d)
M1 ½ × 110 × 85 × sin 56°	✓ 1
A1 3876 m²	✓ 1

Exam tip: When you know two sides and the angle between them, the area is ½ab sin C; you do not need to find any other lengths first.

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A model of the antenna of a space probe is a solid pyramid VABCD. The base ABCD is a horizontal square of side 6 cm. The vertex V is 8 cm vertically above the centre O of the base.

Diagram not to scale

(a)

Calculate the length of the diagonal AC. Give your answer correct to 3 significant figures.

(2)

(b)

Calculate the length of the sloping edge VA. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the angle between VA and the base ABCD. Give your answer correct to 1 decimal place.

(2)

(d)

M is the midpoint of AB. Calculate the angle between the face VAB and the base ABCD. Give your answer correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 √(6² + 6²)	✓ 1
A1 8.49 cm	✓ 1
Part (b)
M1 √(8² + (½ × 8.49)²) = √(64 + 18)	✓ 1
A1 9.06 cm	✓ 1
Part (c)
M1 tan (angle VAO) = 8 ÷ 4.243	✓ 1
A1 62.1°	✓ 1
Part (d)
B1 the angle required is angle VMO, where OM is 3 cm	✓ 1
M1 tan (angle VMO) = 8 ÷ 3	✓ 1
A1 69.4°	✓ 1

Common error: The angle between two planes is measured along lines that are perpendicular to their line of intersection: use the midpoint M of AB, not the corner A.

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Calculator

A grain store is a pyramid VABCD. Its base ABCD is a horizontal rectangle with AB = 24 m and BC = 18 m. The vertex V is vertically above M, the point where the diagonals of the base meet, and VM = 16 m.

Diagram not to scale

(a)

Show that AC = 30 m.

(2)

(b)

Calculate the length VA. Give your answer to 3 significant figures.

(3)

(c)

Calculate the angle between VA and the base. Give your answer to 1 decimal place.

(2)

(d)

Calculate angle AVB. Use the unrounded value of VA. Give your answer to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 AC² = 24² + 18² (= 900)	✓ 1
A1 AC = 30 m	✓ 1
Part (b)
B1 AM = 15 m	✓ 1
M1 VA² = 15² + 16² (= 481)	✓ 1
A1 21.9 m	✓ 1
Part (c)
M1 tan(VAM) = 16 ÷ 15 (or sin = 16 ÷ VA)	✓ 1
A1 46.8°	✓ 1
Part (d)
M1 cosine rule in triangle VAB with VA = VB and AB = 24	✓ 1
M1 cos(AVB) = (481 + 481 − 24²) ÷ (2 × 481)	✓ 1
A1 66.3°	✓ 1

Common error: VM is the height of the pyramid, not the slant edge: the right-angled triangle for VA is VMA, with AM equal to half of the diagonal AC.

+ Add to test✓ In your test384IGMC3-6-11r27-12Pythagoras' theorem and right-angled trigonometry Paper EHard9 marks

Calculator

The displacement, y mm, of the centre of a drum skin from its resting position is modelled by y = 2.5 cos(36t)°, where t is the time in milliseconds and 0 ≤ t ≤ 10.

(a)

Write down the amplitude of the motion and the period of the motion.

(2)

(b)

Find the smallest positive value of t for which y = −1.5. Give your answer to 3 significant figures.

(3)

(c)

Find the two values of t for which y = 1.2. Give each answer to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 amplitude = 2.5 mm	✓ 1
B1 period = 360 ÷ 36 = 10 ms	✓ 1
Part (b)
M1 cos(36t) = −1.5 ÷ 2.5 = −0.6	✓ 1
M1 36t = cos⁻¹(−0.6) = 126.87…°	✓ 1
A1 t = 3.52 ms	✓ 1
Part (c)
M1 cos(36t) = 1.2 ÷ 2.5 = 0.48	✓ 1
A1 t = 1.70 ms	✓ 1
M1 second angle 360° − 61.3…° = 298.7…° (or t = 10 − 1.70)	✓ 1
A1 t = 8.30 ms	✓ 1

Exam tip: Cosine takes each value twice in one cycle, so for a full period always look for the second solution as 360° minus the first.

+ Add to test✓ In your test385IGMC3-6-11r27-14Bearings Paper EHard9 marks

Calculator

Air traffic control tower T tracks two aircraft. Aircraft A is 18 km from T on a bearing of 040°. Aircraft B is 25 km from T on a bearing of 115°.

(a)

Write down angle ATB.

(1)

(b)

Calculate the distance AB. Give your answer to 3 significant figures.

(3)

(c)

Calculate the bearing of B from A. Use the unrounded value of AB. Give your answer to 1 decimal place.

(5)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 115° − 40° = 75°	✓ 1
Part (b)
M1 cosine rule: AB² = 18² + 25² − 2 × 18 × 25 × cos 75°	✓ 1
M1 AB² = 716.0…	✓ 1
A1 26.8 km	✓ 1
Part (c)
M1 sine rule: sin(TAB) ÷ 25 = sin 75° ÷ AB	✓ 1
A1 angle TAB = 64.5° (115.5° is rejected as the other two angles sum to 105°)	✓ 1
B1 bearing of T from A = 220°	✓ 1
M1 bearing of B from A = 220° − 64.5°	✓ 1
A1 155.5°	✓ 1

Common error: Angle TAB is measured at A, so the bearing of B from A starts from the back bearing of T (220°), not from the bearing 040° measured at T.

+ Add to test✓ In your test386IGMC3-6-10r30-12Bearings Paper EHard9 marks

Calculator

A cargo ship carrying coffee beans leaves port P and sails 18 km on a bearing of 035° to a point A. It then sails 25 km on a bearing of 110° to a point B.

(a)

Show that angle PAB is 105°.

(2)

(b)

Calculate the distance PB. Give your answer correct to 3 significant figures.

(3)

(c)

Calculate the bearing of B from P. Give your answer correct to 1 decimal place.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 bearing of P from A is \(035+180=215°\)	✓ 1
B1 \(215-110=105°\)	✓ 1
Part (b)
M1 cosine rule \(PB^2=18^2+25^2-2(18)(25)\cos 105°\)	✓ 1
A1 \(PB^2=1182\) (to 4 s.f.)	✓ 1
A1 34.4 km	✓ 1
Part (c)
M1 sine rule \(\frac{\sin APB}{25}=\frac{\sin 105°}{PB}\)	✓ 1
A1 angle APB = 44.6°	✓ 1
M1 bearing = 035° + angle APB	✓ 1
A1 079.6°	✓ 1

Exam tip: Draw a north line at every point of a bearings problem and mark the angles between the north lines and the path.

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Calculator

A rectangular space-probe test chamber is a cuboid ABCDEFGH. The base ABCD is horizontal with AB = 8 m and BC = 6 m, and the height AE = 5 m, where E is directly above A and G is directly above C.

(a)

Calculate the length of the diagonal AC of the base.

(2)

(b)

Calculate the length of the space diagonal AG. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the angle between AG and the base ABCD. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{8^2+6^2}\)	✓ 1
A1 10 m	✓ 1
Part (b)
M1 \(\sqrt{10^2+5^2}\)	✓ 1
A1 11.2 m	✓ 1
Part (c)
M1 \(\tan^{-1}\left(\frac{5}{10}\right)\)	✓ 1
A1 26.6°	✓ 1

Common error: The angle between a line and a plane is measured to the line's projection onto the plane, here AC, not to an edge of the box.

+ Add to test✓ In your test388IGMC3-6-10r30-14Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EHard8 marks

Calculator

The temperature, \(T\) °C, inside a coffee roasting drum varies with time \(t\) minutes according to \(T=200+15\cos(6t)°\) for \(0\le t\le 60\).

(a)

Write down the amplitude and the period of \(T\).

(2)

(b)

Write down the maximum and minimum values of \(T\).

(2)

(c)

Find the times when \(T=210\). Give your answers correct to 3 significant figures.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 amplitude 15	✓ 1
B1 period \(360\div 6=60\) minutes	✓ 1
Part (b)
B1 maximum 215 °C	✓ 1
B1 minimum 185 °C	✓ 1
Part (c)
M1 \(15\cos(6t)=10\) so \(\cos(6t)=\frac{2}{3}\)	✓ 1
A1 \(6t=48.2°\) so \(t=8.03\) minutes	✓ 1
M1 second solution \(6t=360-48.19...\)	✓ 1
A1 \(t=52.0\) minutes	✓ 1

Exam tip: On a graph of a cosine curve over one period, an equation of the form cos = k has two solutions; find the second using symmetry.

+ Add to test✓ In your test389IGMC3-6-11r33-12Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EHard9 marks

Calculator

In a video game, a temple is modelled as a solid pyramid. Its base ABCD is a horizontal square of side 10 m and its apex V is vertically above the centre M of the base, at a height of 12 m.

(a)

Calculate the length AM, correct to 3 significant figures.

(2)

(b)

Calculate the length of the sloping edge VA, correct to 3 significant figures.

(2)

(c)

Calculate the angle between VA and the base, correct to 1 decimal place.

(2)

(d)

The midpoint of BC is N. Calculate the angle between the face VBC and the base, correct to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 AC = √(10² + 10²) = 14.14... or AM = 5√2	✓ 1
A1 7.07 m	✓ 1
Part (b)
M1 VA² = 12² + 50 (or 12² + 7.07²)	✓ 1
A1 13.9 m	✓ 1
Part (c)
M1 tan x = 12/(5√2) (or 12/7.07...)	✓ 1
A1 59.5°	✓ 1
Part (d)
B1 MN = 5 m	✓ 1
M1 tan y = 12/5	✓ 1
A1 67.4°	✓ 1

Exam tip: In 3D problems, draw the right-angled triangle that contains the angle you want and label its sides with lengths already found.

+ Add to test✓ In your test390IGMC3-6-11r33-13Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EHard9 marks

Calculator

A delivery drone flies between a pharmacy P, a clinic C and a hospital H. The distances are PC = 9.5 km, PH = 14.2 km and CH = 8.0 km.

(a)

Calculate angle CPH, correct to 1 decimal place.

(3)

(b)

The bearing of H from P is 072°. C lies clockwise from PH as seen from P. Calculate the bearing of C from P, correct to 1 decimal place.

(2)

(c)

Calculate the area of triangle PCH, correct to 3 significant figures. Use the unrounded value of angle CPH.

(2)

(d)

Calculate the shortest distance from C to the line PH, correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 cos CPH = (9.5² + 14.2² − 8.0²)/(2 × 9.5 × 14.2)	✓ 1
M1 cos CPH = 0.8446...	✓ 1
A1 32.4°	✓ 1
Part (b)
M1 072° + 32.36...°	✓ 1
A1 104.4°	✓ 1
Part (c)
M1 ½ × 9.5 × 14.2 × sin 32.36...°	✓ 1
A1 36.1 km²	✓ 1
Part (d)
M1 9.5 sin 32.36...° (or 2 × 36.1 ÷ 14.2)	✓ 1
A1 5.08 km	✓ 1

Common error: Store the unrounded angle in your calculator; rounding it to 32.4° before finding the area changes the third significant figure.

+ Add to test✓ In your test391IGMC3-6-10r38-09Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EHard8 marks

Calculator

A lorry leaves a depot O and travels 32 km on a bearing of 065° to a town A. A van leaves the same depot and travels 26 km on a bearing of 140° to a town B.

Diagram not to scale

(a)

Show that angle AOB = 75°.

(1)

(b)

Calculate the distance AB, giving your answer to 3 significant figures.

(3)

(c)

Calculate the bearing of B from A, to the nearest degree. Use the unrounded value of AB.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 140° − 65° = 75°	✓ 1
Part (b)
M1 AB² = 32² + 26² − 2 × 32 × 26 × cos 75° (cosine rule)	✓ 1
A1 AB² = 1269.3... (accept 1269)	✓ 1
A1 AB = 35.6 km	✓ 1
Part (c)
M1 sin(angle OAB) ÷ 26 = sin 75° ÷ 35.63... (sine rule)	✓ 1
A1 angle OAB = 44.8°	✓ 1
M1 bearing of O from A = 065° + 180° = 245°, then subtract angle OAB	✓ 1
A1 200°	✓ 1

Common error: Bearings need a north line at each point; the bearing of B from A is not the same as the bearing of A from B.

+ Add to test✓ In your test392IGMC3-6-10r38-12Pythagoras' theorem and right-angled trigonometry Paper EHard8 marks

Calculator

The roof of a glass atrium is a pyramid with a horizontal square base of side 20 m. The apex V is vertically above the centre of the base at a height of 12 m. Give lengths to 3 significant figures and angles to 1 decimal place.

Diagram not to scale

(a)

Calculate the length of a sloping edge of the pyramid.

(3)

(b)

Calculate the angle between a sloping edge and the base.

(2)

(c)

Calculate the angle between a triangular sloping face and the base.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 diagonal of base = √(20² + 20²), so half-diagonal = 10√2 (= 14.14...)	✓ 1
M1 sloping edge = √((10√2)² + 12²) = √344	✓ 1
A1 18.5 m	✓ 1
Part (b)
M1 tan θ = 12 ÷ (10√2) (or sin θ = 12 ÷ √344)	✓ 1
A1 40.3°	✓ 1
Part (c)
B1 distance from the centre of the base to the midpoint of a base edge = 10 m	✓ 1
M1 tan φ = 12 ÷ 10	✓ 1
A1 50.2°	✓ 1

Exam tip: For the angle between a face and the base, use the line from the centre to the midpoint of a base edge, which is perpendicular to that edge.

+ Add to test✓ In your test393IGMC3-6-10r38-13Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EHard8 marks

No calculator

The height of the tide, h metres, in a harbour at t hours after a high tide is modelled by h = 3.2 + 1.8 cos(30t)°, for 0 ≤ t ≤ 12.

(a)

Write down the maximum and the minimum height of the tide.

(2)

(b)

Write down the time between one high tide and the next.

(1)

(c)

Find the first time after t = 0 when the height of the tide is 4.1 m.

(3)

(d)

Use the symmetry of the cosine graph to find for how many hours of the 12-hour cycle the height of the tide is greater than 4.1 m.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 maximum 5.0 m	✓ 1
B1 minimum 1.4 m	✓ 1
Part (b)
B1 12 hours (period = 360 ÷ 30)	✓ 1
Part (c)
M1 3.2 + 1.8 cos(30t) = 4.1, so cos(30t) = 0.5	✓ 1
A1 30t = 60	✓ 1
A1 t = 2 hours	✓ 1
Part (d)
M1 the other solution in the cycle is t = 10 (30t = 300), and the height is above 4.1 m for 0 ≤ t < 2 and 10 < t ≤ 12	✓ 1
A1 4 hours	✓ 1

Common error: Cosine equations have a second solution in each cycle; use the symmetry of the graph to find it instead of stopping at the calculator value.

+ Add to test✓ In your test394IGMC3-6-10r40-13Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EHard10 marks

Calculator

A model of a mountain is a pyramid VABCD. The base ABCD is a horizontal rectangle with AB = 80 m and BC = 60 m. The apex V is vertically above the centre of the base and is 65 m above the base. Use unrounded values in your working.

Diagram not to scale

(a)

Show that the diagonal AC is 100 m.

(2)

(b)

Calculate the length VA. Give your answer to 3 significant figures.

(3)

(c)

Calculate the angle between VA and the base. Give your answer to 1 decimal place.

(2)

(d)

Calculate angle AVB. Give your answer to 1 decimal place.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{80^2+60^2}\)	✓ 1
A1 100	✓ 1
Part (b)
M1 half the diagonal is 50 m	✓ 1
M1 \(VA^2=65^2+50^2=6725\)	✓ 1
A1 82.0 m	✓ 1
Part (c)
M1 \(\tan x=\dfrac{65}{50}\)	✓ 1
A1 52.4°	✓ 1
Part (d)
M1 \(80^2=6725+6725-2(6725)\cos(AVB)\)	✓ 1
M1 \(\cos(AVB)=\dfrac{13450-6400}{13450}\)	✓ 1
A1 58.4°	✓ 1

Common error: VA and VB are equal in length, so use the cosine rule in the isosceles triangle AVB, not a right-angled triangle.

+ Add to test✓ In your test395IGMC3-6-10r40-14Pythagoras' theorem and right-angled trigonometry Paper EHard9 marks

Calculator

Three telecoms towers A, B and C are on level ground. AB = 14 km, BC = 9 km and AC = 18 km. Tower B is due east of tower A and tower C is north of the line AB.

Diagram not to scale

(a)

Use the cosine rule to calculate angle ABC. Give your answer to 1 decimal place.

(3)

(b)

Calculate the area of triangle ABC. Give your answer to 3 significant figures.

(2)

(c)

Use the sine rule to calculate angle CAB. Give your answer to 1 decimal place.

(3)

(d)

Find the bearing of C from A. Give your answer to 1 decimal place.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\cos B=\dfrac{14^2+9^2-18^2}{2(14)(9)}\)	✓ 1
A1 \(\cos B=-\dfrac{47}{252}\) (accept −0.187)	✓ 1
A1 100.7°	✓ 1
Part (b)
M1 \(\dfrac12\times14\times9\times\sin B\)	✓ 1
A1 61.9 km²	✓ 1
Part (c)
M1 \(\dfrac{\sin A}{9}=\dfrac{\sin B}{18}\)	✓ 1
M1 \(A=\sin^{-1}\left(\dfrac{9\sin B}{18}\right)\)	✓ 1
A1 29.4°	✓ 1
Part (d)
B1 060.6° (ECF from (c): \(90°-\) their angle CAB)	✓ 1

Exam tip: The sine rule is safe here because angle A lies opposite the shortest side, so it must be acute.

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Calculator

Points A, B and F lie in a straight line on level ground, with B between A and F. The point F is the foot of a vertical pylon FT of a bridge. The height of the pylon is h metres, AB = 30 m and BF = x metres. The angle of elevation of T from A is 25° and the angle of elevation of T from B is 40°.

Diagram not to scale

(a)

Write down two expressions for h, one in terms of x and 40°, and one in terms of x and 25°.

(2)

(b)

Find the value of x. Give your answer correct to 3 significant figures.

(3)

(c)

Calculate the height of the pylon. Give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 \(h=x\tan 40^\circ\)	✓ 1
B1 \(h=(x+30)\tan 25^\circ\)	✓ 1
Part (b)
M1 equate: \(x\tan40^\circ=(x+30)\tan25^\circ\)	✓ 1
M1 collect terms in x: \(x(\tan40^\circ-\tan25^\circ)=30\tan25^\circ\)	✓ 1
A1 37.5	✓ 1
Part (c)
M1 \(h=x\tan40^\circ\) with their x (ECF from (b); accept 37.5 or the unrounded value)	✓ 1
A1 31.5 m	✓ 1

Exam tip: With two angles of elevation from the same line, use one unknown distance and write the same height in two ways. Equating the two expressions gives an equation you can solve.

+ Add to test✓ In your test397IGMC3-6-11r40-10Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EHard8 marks

Calculator

A coffee warehouse is a cuboid ABCDEFGH. The floor ABCD is a rectangle with AB = 30 m and BC = 16 m. The vertical edges AE, BF, CG and DH each have length 9 m, so that E is directly above A and G is directly above C.

Diagram not to scale

(a)

Calculate the length of the floor diagonal AC.

(2)

(b)

Calculate the length of the diagonal AG. Give your answer correct to 3 significant figures.

(2)

(c)

Calculate the angle between AG and the floor ABCD. Give your answer correct to 1 decimal place.

(2)

(d)

The point M is the midpoint of CG. A sensor cable runs in a straight line from A to M. Calculate the angle between AM and the floor. Give your answer correct to 1 decimal place.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{30^2+16^2}\)	✓ 1
A1 34 m	✓ 1
Part (b)
M1 \(\sqrt{34^2+9^2}\) or \(\sqrt{1237}\)	✓ 1
A1 35.2 m	✓ 1
Part (c)
M1 \(\tan(\text{GAC})=\frac{9}{34}\) or \(\sin(\text{GAC})=\frac{9}{35.17}\)	✓ 1
A1 14.8°	✓ 1
Part (d)
M1 \(\tan(\text{MAC})=\frac{4.5}{34}\)	✓ 1
A1 7.5°	✓ 1

Common error: The angle between a line and a plane is measured to the line's shadow on that plane, here AC, not to an edge of the box such as AB.

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Two research boats leave harbour H at the same time. Boat A sails 16 km on a bearing of 062°. Boat B sails 23 km on a bearing of 125°. Both boats sail in straight lines.

Diagram not to scale

(a)

Write down the size of angle AHB.

(1)

(b)

Calculate the distance AB. Give your answer correct to 3 significant figures.

(3)

(c)

Calculate angle HBA. Give your answer correct to 1 decimal place. (Angle HBA is acute because it is opposite the shortest side.)

(2)

(d)

Find the bearing of A from B, to the nearest degree.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 63°	✓ 1
Part (b)
M1 \(AB^2=16^2+23^2-2(16)(23)\cos63^\circ\)	✓ 1
M1 \(AB^2=450.9\) (accept 451) or evaluate square root	✓ 1
A1 21.2 km	✓ 1
Part (c)
M1 \(\frac{\sin(\text{HBA})}{16}=\frac{\sin63^\circ}{21.23}\) (accept 21.2)	✓ 1
A1 42.2°	✓ 1
Part (d)
B1 bearing of H from B is 305°	✓ 1
M1 305° + their angle HBA (ECF from (c))	✓ 1
A1 347°	✓ 1

Exam tip: Draw a north line at each point and mark the given bearings. When using the sine rule to find an angle, say whether it is acute or obtuse and why.

+ Add to test✓ In your test399IGMC3-6-10r45-13Pythagoras' theorem and right-angled trigonometry Paper EHard9 marks

Calculator

The height, h cm, of a cyclist's pedal above the ground is modelled by h = 32 + 17 cos(240t)°, where t is the time in seconds and the angle 240t is measured in degrees.

(a)

Write down the maximum and minimum heights of the pedal.

(2)

(b)

Find the time for one complete turn of the pedal.

(1)

(c)

Find the two times in the first turn, 0 ≤ t ≤ 1.5, when the pedal is 40 cm above the ground. Give your answers correct to 3 significant figures.

(4)

(d)

For how long in each turn is the pedal more than 40 cm above the ground? Use unrounded values from (c) and give your answer correct to 3 significant figures.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 maximum 49 cm	✓ 1
B1 minimum 15 cm	✓ 1
Part (b)
B1 360 ÷ 240 = 1.5 s	✓ 1
Part (c)
M1 32 + 17 cos(240t) = 40, so cos(240t) = 8/17	✓ 1
M1 240t = cos⁻¹(8/17) = 61.9...°	✓ 1
A1 t = 0.258 s	✓ 1
A1 second solution 240t = 360° − 61.9...°, so t = 1.24 s	✓ 1
Part (d)
M1 the pedal is above 40 cm for 0 ≤ t < 0.258 and again for 1.242 < t ≤ 1.5, so the total time is 2 × 0.258 (or 1.5 − (1.242 − 0.258))	✓ 1
A1 0.516 s	✓ 1

Common error: A cosine equation has a second solution in the full cycle: use 360° minus the calculator value.

+ Add to test✓ In your test400IGMC3-6-10r45-14Bearings Paper EHard10 marks

Calculator

Anna and Ben leave a bridge O at the same time. Anna cycles on a bearing of 035° at a steady 18 km/h. Ben cycles on a bearing of 128° at a steady 24 km/h. After 2.5 hours Anna is at A and Ben is at B.

Diagram not to scale

(a)

Calculate the distances OA and OB.

(2)

(b)

Show that angle AOB = 93°.

(1)

(c)

Calculate the distance AB. Give your answer correct to 3 significant figures.

(3)

(d)

Calculate the bearing of B from A. Give your answer correct to 1 decimal place.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
B1 OA = 18 × 2.5 = 45 km	✓ 1
B1 OB = 24 × 2.5 = 60 km	✓ 1
Part (b)
B1 128° − 035° = 93°	✓ 1
Part (c)
M1 AB² = 45² + 60² − 2 × 45 × 60 × cos 93°	✓ 1
A1 AB² = 5908 (5907.6...)	✓ 1
A1 AB = 76.9 km	✓ 1
Part (d)
M1 sin(OAB) ÷ 60 = sin 93° ÷ 76.9	✓ 1
A1 angle OAB = 51.2°	✓ 1
M1 the bearing of O from A is 035° + 180° = 215°	✓ 1
A1 215° − 51.2° = 163.8°	✓ 1

Exam tip: Convert speed and time into distances first; then the two legs and the angle between their bearings give a triangle with two sides and the included angle.

+ Add to test✓ In your test401IGMC3-6-12r45-09Bearings Paper EHard7 marks

Calculator

Two drones leave a hospital helipad H at the same time. Drone P flies on a bearing of 035° for 6.2 km. Drone Q flies on a bearing of 118° for 8.4 km.

Diagram not to scale

(a)

Calculate the distance PQ. Give your answer to 3 significant figures.

(3)

(b)

Calculate the bearing of Q from P. Give your answer to 1 decimal place.

(4)

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Marking point	Mark	Notes
Part (a)
B1 angle PHQ = 118° − 35° = 83°	✓ 1
M1 PQ² = 6.2² + 8.4² − 2 × 6.2 × 8.4 × cos 83°	✓ 1
A1 9.81 km	✓ 1
Part (b)
M1 sin(angle HPQ) ÷ 8.4 = sin 83° ÷ 9.81	✓ 1
A1 angle HPQ = 58.2° (acute, since HQ is not the longest side)	✓ 1
M1 bearing of H from P is 215°	✓ 1
A1 bearing of Q from P = 215° − 58.2° = 156.8°	✓ 1

Common error: When using the sine rule to find an angle, check whether it could be obtuse; here the side opposite is not the longest, so the angle is acute.

+ Add to test✓ In your test402IGMC3-6-11r46-09Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EHard6 marks

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An atrium in a museum is a cuboid ABCDEFGH. The floor is the rectangle ABCD with AB = 12 m and BC = 9 m. The vertical edges AE, BF, CG and DH each have length 5 m. A cable runs in a straight line from corner A on the floor to corner G on the ceiling, where G is directly above C.

Diagram not to scale

(a)

Calculate the length of the diagonal AC of the floor.

(2)

(b)

Calculate the length of the cable AG. Give your answer to 3 significant figures.

(2)

(c)

Calculate the angle between the cable AG and the floor. Give your answer to 1 decimal place.

(2)

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Marking point	Mark	Notes
Part (a)
M1 \(\sqrt{12^2+9^2}\)	✓ 1
A1 15 m	✓ 1
Part (b)
M1 \(\sqrt{15^2+5^2}\) (ECF from (a))	✓ 1
A1 15.8 m	✓ 1
Part (c)
M1 \(\tan\theta=\frac{5}{15}\) (angle GAC)	✓ 1
A1 18.4°	✓ 1

Exam tip: For angles in 3D shapes, draw the right-angled triangle that contains the line and its projection onto the plane.

+ Add to test✓ In your test403IGMC3-6-11r46-14Bearings Paper EHard8 marks

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Two ships leave port P at the same time. Ship A sails 15 km on a bearing of 050° and ship B sails 22 km on a bearing of 125°.

Diagram not to scale

(a)

Write down the size of angle APB.

(1)

(b)

Calculate the distance AB. Give your answer to 3 significant figures.

(3)

(c)

Calculate the bearing of B from A, giving your answer to 1 decimal place. Use the unrounded value of AB.

(4)

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Marking point	Mark	Notes
Part (a)
B1 75°	✓ 1
Part (b)
M1 cosine rule \(15^2+22^2-2\times15\times22\cos75^\circ\)	✓ 1
M1 evaluates (538.2) and takes the square root	✓ 1
A1 23.2 km	✓ 1
Part (c)
M1 uses the sine rule or cosine rule to find angle PAB (angle PAB is acute because it is not opposite the longest side)	✓ 1
A1 66.35° (accept 66.4°)	✓ 1
M1 bearing of P from A is 230°, so bearing of B is \(230-66.35\)	✓ 1
A1 163.6° (accept 163.7° if 66.35° is used)	✓ 1

Common error: When using the sine rule to find an angle, check whether the angle could be obtuse; here angle PAB is acute because AB is the longest side.