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Cambridge IGCSE Mathematics 0580 · Core and Extended · Topic 6

Trigonometry

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Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 15 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →

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+ Add to test✓ In your test1IGMC-046-01-01Bearings Paper CEasy4 marks

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°.

(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°.

+ Add to test✓ In your test2IGMC-046-01-02Bearings Paper CEasy5 marks

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.

(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.

+ Add to test✓ In your test3IGMC-046-01-03Bearings Paper CEasy5 marks

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°.

(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.

+ Add to test✓ In your test4IGMC-046-01-04Bearings Paper CEasy4 marks

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'.

+ Add to test✓ In your test5IGMC-046-01-06Bearings Paper CEasy5 marks

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.

(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.

+ Add to test✓ In your test6IGMC-046-00-01Pythagoras' theorem and right-angled trigonometry Paper CEasy5 marks

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.

(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°.

+ Add to test✓ In your test7IGMC-046-00-02Pythagoras' theorem and right-angled trigonometry Paper CEasy4 marks

No calculator

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

(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.

+ Add to test✓ In your test8IGMC-046-00-03Pythagoras' theorem and right-angled trigonometry Paper CEasy6 marks

Calculator

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

(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.

+ Add to test✓ In your test9IGMC-046-00-05Pythagoras' theorem and right-angled trigonometry Paper CEasy4 marks

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.

+ Add to test✓ In your test10IGMC-046-00-11Pythagoras' theorem and right-angled trigonometry Paper CEasy5 marks

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.

+ Add to test✓ In your test11IGMC-046-01-05Bearings Paper CMedium6 marks

No calculator

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

(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 test12IGMC-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°.

(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 test13IGMC-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.

(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 test14IGMC-046-00-04Pythagoras' theorem and right-angled trigonometry 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.

(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 test15IGMC-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.

(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 test16IGMC-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°.

(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 test17IGMC-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.

(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 test18IGMC-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°.

(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 test19IGMC-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.

(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.

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Calculator

In triangle PQR, PQ = 7.2 cm, PR = 5.8 cm and angle QPR = 112°.

(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 test21IGMC-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.

(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 test22IGMC-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 test23IGMC-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 test24IGMC-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.

(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 test25IGMC-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.

(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 test26IGMC-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.

(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 test27IGMC-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° = \(\frac{\sqrt{3}}{2}\).

(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 test28IGMC-046-02-06Sine rule, cosine rule, trigonometric graphs and 3D problems 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.

(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 test29IGMC-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.

(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 test30IGMC-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.

(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 test31IGMC-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° = \(\frac{\sqrt{3}}{2}\).

(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 test32IGMC-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 test33IGMC-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.

(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 test34IGMC-046-01-12Bearings Paper EHard7 marks

Calculator

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.

(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 test35IGMC-046-01-13Bearings Paper EHard8 marks

Calculator

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°.

(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 test36IGMC-046-01-14Bearings Paper EHard8 marks

Calculator

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.

(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 test37IGMC-046-00-12Pythagoras' theorem and right-angled trigonometry Paper EHard7 marks

Calculator

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.

(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 test38IGMC-046-00-13Pythagoras' theorem and right-angled trigonometry Paper EHard7 marks

Calculator

A right-angled triangle has sides of length x cm, (x + 7) cm and (x + 8) cm, where (x + 8) cm is the hypotenuse.

(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 test39IGMC-046-00-14Pythagoras' theorem and right-angled trigonometry Paper EHard5 marks

Calculator

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.

(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.

+ Add to test✓ In your test40IGMC-046-02-12Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EHard9 marks

Calculator

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°.

(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 test41IGMC-046-02-13Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EHard8 marks

Calculator

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°.

(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 test42IGMC-046-02-14Sine rule, cosine rule, trigonometric graphs and 3D problems Paper EHard8 marks

No calculator

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° = \(\frac{\sqrt{3}}{2}\).

(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).