311 free, high-quality past-paper-style questions on Geometry and trigonometry for IB Maths AA SL, each with a full mark scheme. Covers: Three-dimensional geometry; Right-angled and non-right-angled trigonometry; Radian measure, arcs and sectors; and more. For when you have finished the past papers, or want to practise one topic at a time.Topic 3 at Higher Level →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.2,368 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1AASL3-q00013.1 Three-dimensional geometryPaper 1Easy7 marks
The points \(A\) and \(B\) have coordinates \((-1,\,4,\,2)\) and \((3,\,-2,\,14)\) respectively.
(a)
Find the distance \(AB\).
(2)
(b)
Find the coordinates of \(M\), the midpoint of \([AB]\).
(2)
(c)
The point \(C\) has coordinates \((3,\,k,\,2)\), where \(k\in\mathbb{R}\). Given that \(CM=7\), find the possible values of \(k\).
2AASL3-q00023.1 Three-dimensional geometryPaper 1Easy7 marks
Diagram not to scale
A toy is made from a solid right circular cone joined to a solid hemisphere. The cone and the hemisphere have a common circular face of radius \(3\) cm, and the height of the cone is \(8\) cm, as shown in the diagram.
(a)
Find the volume of the toy. Give your answer in the form \(k\pi\text{ cm}^{3}\), where \(k\in\mathbb{Z}\).
4AASL3-q00043.1 Three-dimensional geometryPaper 1Easy6 marks
A cylindrical container with internal radius \(6\) cm stands on a horizontal table and contains water. Metal balls, each a sphere of radius \(1.5\) cm, are dropped into the container. Every ball sinks and is completely covered by the water, and no water overflows.
(a)
Find the exact volume of one ball.
(2)
(b)
Twenty balls are dropped into the container. Show that the water level rises by \(2.5\) cm.
(2)
(c)
The container is emptied of balls, refilled to its original level, and \(n\) balls are dropped in. Find the least value of \(n\) for which the water level rises by at least \(3.7\) cm.
19IBMA2S-048-05-013.6 Trigonometric identitiesPaper 1Easy5 marks
No calculator
A robotic arm on a car assembly line makes an angle \(\theta\) with the horizontal, where \(\frac{\pi}{2}<\theta<\pi\) and \(\sin\theta=\frac{5}{13}\).
(a)
Find the exact value of \(\cos\theta\).
(2)
(b)
Find the exact value of \(\tan\theta\).
(1)
(c)
Find the exact value of \(\sin 2\theta\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(\sin^2\theta+\cos^2\theta=1\), giving \(\cos^2\theta=\frac{144}{169}\)
✓ 1
A1 \(\cos\theta=-\frac{12}{13}\), negative because \(\theta\) is obtuse
Common error: Taking the positive square root for \(\cos\theta\) loses the final marks. Always check which quadrant the angle lies in before choosing the sign.
20IBMA2S-048-05-023.6 Trigonometric identitiesPaper 1Easy4 marks
No calculator
In an electronics lab, the squared output of a mixing circuit is modelled by \(S(x)=(\sin x+\cos x)^2\).
(a)
Show that \(S(x)=1+\sin 2x\).
(2)
(b)
Hence find the exact value of \(\left(\sin\frac{\pi}{12}+\cos\frac{\pi}{12}\right)^2\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 expands to \(\sin^2x+2\sin x\cos x+\cos^2x\)
✓ 1
A1 uses \(\sin^2x+\cos^2x=1\) and \(2\sin x\cos x=\sin 2x\) to obtain \(1+\sin 2x\)
✓ 1
Part (b)
M1 substitutes \(x=\frac{\pi}{12}\) to obtain \(1+\sin\frac{\pi}{6}\)
✓ 1
A1 \(\frac{3}{2}\)
✓ 1
Exam tip: When a part says "hence", use the result just proved. Here doubling \(\frac{\pi}{12}\) turns an awkward angle into a standard one.
21IBMA2S-048-05-043.6 Trigonometric identitiesPaper 1Easy4 marks
No calculator
An architect needs the exact gradient of a roof pitched at \(\frac{\pi}{8}\) radians.
(a)
Show that \(\frac{1-\cos 2x}{\sin 2x}=\tan x\) for \(\sin 2x\neq 0\).
(2)
(b)
Hence find the exact value of \(\tan\frac{\pi}{8}\), giving your answer in the form \(a+b\sqrt{2}\) where \(a,b\in\mathbb{Z}\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes \(\cos 2x=1-2\sin^2x\) and \(\sin 2x=2\sin x\cos x\) to get \(\frac{2\sin^2x}{2\sin x\cos x}\)
Exam tip: Choose the version of \(\cos 2x\) that cancels the 1: here \(1-2\sin^2x\) makes the numerator a single term.
22IBMA2S-048-07-013.4 Radian measure, arcs and sectorsPaper 1Easy4 marks
No calculator
A windscreen wiper on a bus has a blade arm of length 48 cm, pivoted at one end. As the arm rotates, its tip moves along a circular arc of length 40π cm.
Diagram not to scale
(a)
Find, in radians, the angle through which the wiper arm rotates.
(2)
(b)
Find the exact area of the sector swept out by the wiper arm.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses s = rθ: 40π = 48θ
✓ 1
A1 θ = 5π/6
✓ 1
Part (b)
M1 uses A = ½r²θ = ½ × 48² × 5π/6, or A = ½rs = ½ × 48 × 40π
✓ 1
A1 960π cm²
✓ 1
Exam tip: If you know the radius and the arc length, A = ½rs gives the sector area directly, without finding θ first.
23IBMA2S-048-07-033.4 Radian measure, arcs and sectorsPaper 1Easy4 marks
No calculator
When a folding hand fan is fully open, it forms a sector of a circle with a central angle of 1.5 radians. A decorative ribbon is stitched around the entire perimeter of the open fan, along both straight edges and the curved edge. The ribbon has a total length of 35 cm.
Diagram not to scale
(a)
Find the radius of the fan.
(2)
(b)
Find the area of the open fan.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 perimeter equation 2r + 1.5r = 35
✓ 1
A1 r = 10 cm
✓ 1
Part (b)
M1 A = ½ × 10² × 1.5
✓ 1
A1 75 cm²
✓ 1
Common error: The perimeter of a sector includes both radii. Writing only rθ = 35 leaves out the two straight edges.
24IBMA2S-048-02-013.5 The unit circle and the ambiguous casePaper 1Easy5 marks
No calculator
The tip of a rotating radar beam is modelled as a point P on the unit circle, where θ is the angle measured anticlockwise from the positive x-axis to OP. At a certain instant, π < θ < 3π/2 and cos θ = −√5/3.
A1 sin θ = −2/3, negative because θ is in the third quadrant
✓ 1
Part (b)
M1 uses tan θ = sin θ / cos θ = (−2/3) ÷ (−√5/3)
✓ 1
A1 tan θ = 2/√5 = 2√5/5
✓ 1
Part (c)
A1 cos(π − θ) = −cos θ = √5/3
✓ 1
Exam tip: Decide the sign from the quadrant before you take the square root: in the third quadrant both sine and cosine are negative, so tangent is positive.
25IBMA2S-048-02-043.6 Trigonometric identitiesPaper 1Easy4 marks
No calculator
The phase angle x radians of an alternating signal satisfies √2 sin x = −1, where −π ≤ x ≤ π.
(a)
Find all possible values of x.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sin x = −1/√2
✓ 1
A1 reference angle π/4
✓ 1
A1 x = −π/4
✓ 1
A1 x = −3π/4
✓ 1
Common error: Giving 5π/4 and 7π/4 ignores the stated interval; on −π ≤ x ≤ π the negative-sine solutions lie below the x-axis as −π/4 and −3π/4.
26IBMA2S-048-02-113.5 The unit circle and the ambiguous casePaper 1Easy5 marks
No calculator
Two security cameras A and B move on a circular track of radius 1 m centred at the origin O. Their positions are given by angles measured anticlockwise from the positive x-axis: camera A is at angle 5π/6 and camera B is at angle −π/2.
(a)
Write down the exact coordinates of A and of B.
(2)
(b)
Find the exact straight-line distance AB.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 A(−√3/2, 1/2)
✓ 1
A1 B(0, −1)
✓ 1
Part (b)
M1 distance formula: AB2 = (√3/2)2 + (3/2)2, or cosine rule with angle AOB = 2π/3
✓ 1
A1 AB2 = 3/4 + 9/4 = 3
✓ 1
A1 AB = √3 m
✓ 1
Exam tip: Any point on the unit circle at angle θ has coordinates (cos θ, sin θ); a negative angle simply means measuring clockwise.
27IBMA2S-048-03-013.7 Circular functions and their graphsPaper 1Easy5 marks
No calculator
The blood pressure, P mmHg, in a patient's artery is modelled by \(P(t) = 100 + 20\sin\left(\frac{5\pi t}{2}\right)\), where t is the time in seconds after monitoring begins.
(a)
Find the period of the model.
(1)
(b)
Find the two values of t in the first period, \(0 \le t \le \frac{4}{5}\), for which P = 90. Give your answers as exact fractions.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 period \(= \frac{2\pi}{5\pi/2} = \frac{4}{5}\) s
A1 \(t = \frac{7}{15}\) s and \(t = \frac{11}{15}\) s
✓ 1
Common error: A negative sine value means the angles lie in the third and fourth quadrants. Writing \(-\frac{\pi}{6}\) gives an angle outside the required range. Use \(\pi + \frac{\pi}{6}\) and \(2\pi - \frac{\pi}{6}\) instead.
28IBMA2S-048-03-073.8 Trigonometric equationsPaper 1Easy4 marks
No calculator
Solve the equation \(\tan 3x = -1\) for \(0 \le x \le \pi\).
Exam tip: Scale the interval before solving. Because \(3x\) runs from 0 to \(3\pi\), the tangent repeats every π and you need every value of the form \(\frac{3\pi}{4} + k\pi\) in that range.
29IBMA2S-048-00-013.1 Three-dimensional geometryPaper 1Easy6 marks
No calculator
A survey drone is tracked using coordinates in metres, with the launch pad at the origin and the z-axis pointing vertically upwards. The drone hovers at A(2, −1, 17) and later at B(8, 3, 5).
(a)
A ground beacon records that the point P(p, 7, 17) is exactly 10 m from A. Find the possible values of p.
(2)
(b)
Find the coordinates of the midpoint of [AB].
(2)
(c)
Find the distance AB.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sets up (p − 2)2 + 82 + 02 = 100
✓ 1
A1 p = 8 or p = −4
✓ 1
Part (b)
M1 averages the coordinates, e.g. ((2+8)/2, (−1+3)/2, (17+5)/2)
✓ 1
A1 (5, 1, 11)
✓ 1
Part (c)
M1 √(62 + 42 + (−12)2) = √196
✓ 1
A1 AB = 14 m
✓ 1
Common error: Squaring a negative difference such as −12 must give +144. Dropping the sign before squaring is fine, but never subtract the square.
30IBMA2S-048-00-033.1 Three-dimensional geometryPaper 1Easy7 marks
No calculator
A solid bronze hemisphere of radius 6 cm is melted down and recast, with no loss of metal, into a solid right circular cone for a trophy. The cone has base radius 6 cm.
(a)
Find the volume of the hemisphere, giving your answer in terms of π.
(2)
(b)
Find the height of the cone.
(2)
(c)
Find the exact curved surface area of the cone.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (2/3)π(6)3
✓ 1
A1 144π cm3
✓ 1
Part (b)
M1 (1/3)π(6)2h = 144π
✓ 1
A1 h = 12 cm
✓ 1
Part (c)
M1 slant height l = √(62 + 122) = √180 = 6√5
✓ 1
M1 uses πrl = π(6)(6√5)
✓ 1
A1 36√5π cm2
✓ 1
Common error: The curved surface area of a cone uses the slant height, not the vertical height. Using 12 instead of 6√5 loses the accuracy mark.
31IBMA2S-048-01-013.8 Trigonometric equationsPaper 1Easy5 marks
No calculator
The function \(f(x)=a\cos(bx)+d\), where \(a>0\) and \(b>0\), has a maximum point at \((0,\,7)\). The next minimum point to the right of the \(y\)-axis is at \(\left(\frac{\pi}{3},\,-1\right)\).
(a)
Find the value of \(a\).
(2)
(b)
Find the value of \(d\).
(1)
(c)
Find the value of \(b\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use \(a=\frac{\text{max}-\text{min}}{2}=\frac{7-(-1)}{2}\)
✓ 1
A1 \(a=4\)
✓ 1
Part (b)
A1 \(d=\frac{7+(-1)}{2}=3\)
✓ 1
Part (c)
M1 maximum to next minimum is half a period, so period \(=\frac{2\pi}{3}\)
✓ 1
A1 \(b=\frac{2\pi}{2\pi/3}=3\)
✓ 1
Common error: The horizontal distance from a maximum to the next minimum is half a period, not a full period. Using π/3 as the full period gives b = 6, which is wrong.
32IBMA2S-048-01-033.7 Circular functions and their graphsPaper 1Easy7 marks
No calculator
The graph of \(y=\sin x\) is transformed by the following sequence: • a vertical stretch with scale factor 3; • a horizontal stretch with scale factor \(\frac{1}{2}\); • a translation by the vector \(\begin{pmatrix}\pi/4\\1\end{pmatrix}\). The resulting graph is \(y=g(x)\).
(a)
Write \(g(x)\) in the form \(a\sin\big(b(x-c)\big)+d\).
(3)
(b)
Find the range of \(g\).
(2)
(c)
Find the exact value of \(g\left(\frac{\pi}{3}\right)\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(a=3\)
✓ 1
A1 \(b=2\)
✓ 1
A1 \(c=\frac{\pi}{4}\) and \(d=1\), giving \(g(x)=3\sin\left(2\left(x-\frac{\pi}{4}\right)\right)+1\)
Common error: A horizontal stretch with scale factor 1/2 multiplies x by 2 inside the function. The translation then replaces x with (x − π/4), so the 2 multiplies the whole bracket, not just x.
33IBMA2S-048-06-023.2 Right-angled and non-right-angled trigonometryPaper 1Easy4 marks
No calculator
A wheelchair ramp is built in the shape of a right-angled triangle. The sloping surface makes an angle of 30° with the horizontal ground, and the horizontal length of the ramp is \(4\sqrt{3}\) m.
Diagram not to scale
(a)
Find the vertical height of the ramp.
(2)
(b)
Find the length of the sloping surface of the ramp.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(h = 4\sqrt{3}\tan 30°\)
✓ 1
A1 \(h = 4\sqrt{3}\times\frac{1}{\sqrt{3}} = 4\) m
✓ 1
Part (b)
M1 \(L = \frac{4\sqrt{3}}{\cos 30°}\) or \(L = \frac{4}{\sin 30°}\) or Pythagoras
✓ 1
A1 \(L = 8\) m
✓ 1
Exam tip: Learn the exact values \(\sin 30° = \frac{1}{2}\), \(\cos 30° = \frac{\sqrt{3}}{2}\) and \(\tan 30° = \frac{1}{\sqrt{3}}\). A quick Pythagoras check, \(4^2 + (4\sqrt{3})^2 = 64 = 8^2\), confirms both answers.
34IBMA2S-048-04-023.4 Radian measure, arcs and sectorsPaper 1Easy5 marks
No calculator
A triangular shade sail is fixed to a garden wall. Two edges of the sail have lengths 6 m and 8 m, and the angle between these two edges is 150°.
Diagram not to scale
(a)
Find the area of the sail.
(2)
(b)
Find the exact length of the third edge of the sail.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 Area \(=\frac{1}{2}\times 6\times 8\times\sin 150°\)
Common error: cos 150° is negative, so the term −2ab cos C becomes positive. Writing c² = 100 − 48√3 gives an edge shorter than expected for an obtuse angle, which should signal a sign slip.
35IBMA3S-3-17r6-043.6 Trigonometric identitiesPaper 1Easy5 marks
No calculator
The angle \(\theta\) is obtuse and \(\sin\theta=\dfrac35\).
(a)
Find the exact value of \(\cos\theta\).
(2)
(b)
Find the exact value of \(\sin 2\theta\).
(2)
(c)
Find the exact value of \(\cos 2\theta\).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(\cos^2\theta=1-\sin^2\theta\) (or a right-angled triangle with sides 3, 4, 5)
✓ 1
A1 \(\cos\theta=-\frac45\) (negative since obtuse)
Common error: Check the sign of cos θ from the quadrant before using a double-angle formula.
36IBMA3S-3-17r8-043.6 Trigonometric identitiesPaper 1Easy4 marks
No calculator
The vertical ground displacement \(d\) mm recorded near a volcano is modelled by \(d=4\sin t\), where \(t\) is in radians. A scientist wants the times in one cycle when \(d=2\sqrt{3}\).
(a)
Solve \(4\sin t=2\sqrt{3}\) for \(0\le t\le 2\pi\), giving exact answers.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\sin t=\frac{\sqrt3}{2}\)
✓ 1
A1 \(t=\frac{\pi}{3}\)
✓ 1
M1 second solution using \(\pi-\frac{\pi}{3}\)
✓ 1
A1 \(t=\frac{2\pi}{3}\)
✓ 1
Exam tip: In a restricted domain, find the principal value first and then use symmetry of the sine graph to find the other solution.
37IBMA3S-3-16r18-023.6 Trigonometric identitiesPaper 1Easy4 marks
No calculator
The angle \(\theta\) is acute and \(\sin\theta=\frac{2}{3}\).
(a)
Find the exact value of \(\cos\theta\).
(2)
(b)
Find the exact value of \(\sin2\theta\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(\sin^2\theta+\cos^2\theta=1\) to get \(\cos^2\theta=\frac{5}{9}\)
✓ 1
A1 \(\cos\theta=\frac{\sqrt5}{3}\) (positive since \(\theta\) is acute)
✓ 1
Part (b)
M1 use of \(\sin2\theta=2\sin\theta\cos\theta\)
✓ 1
A1 \(\sin2\theta=2\times\frac{2}{3}\times\frac{\sqrt5}{3}=\frac{4\sqrt5}{9}\) (ECF from (a))
✓ 1
Exam tip: Check the sign of the cosine from the quadrant stated in the question before you finalise it.
38IBMA3S-3-16r18-093.6 Trigonometric identitiesPaper 1Easy5 marks
No calculator
Do not use a calculator in this question.
(a)
Convert \(135^\circ\) into radians, giving your answer as a multiple of \(\pi\).
(2)
(b)
Find the exact value of \(\sin\frac{2\pi}{3}+\cos\frac{\pi}{3}\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(135\times\frac{\pi}{180}\)
✓ 1
A1 \(\frac{3\pi}{4}\)
✓ 1
Part (b)
A1 \(\sin\frac{2\pi}{3}=\frac{\sqrt3}{2}\)
✓ 1
A1 \(\cos\frac{\pi}{3}=\frac{1}{2}\)
✓ 1
A1 sum \(=\frac{1+\sqrt3}{2}\)
✓ 1
Exam tip: Learn the exact values for \(\frac{\pi}{6}\), \(\frac{\pi}{4}\) and \(\frac{\pi}{3}\) and use symmetry for the other angles.
39IBMA3S-3-17r24-023.4 Radian measure, arcs and sectorsPaper 1Easy5 marks
No calculator
A theatre spotlight lights a sector-shaped region of the stage floor. The sector has radius 6 m and the angle at the centre is \(\frac{\pi}{3}\) radians.
Diagram not to scale
(a)
Find the length of the curved edge of the lit region, in terms of π.
(2)
(b)
Find the area of the lit region, in terms of π.
(2)
(c)
Write down the perimeter of the lit region, in the form a + bπ.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of l = rθ = 6 × π/3
✓ 1
A1 2π m
✓ 1
Part (b)
M1 use of A = ½r2θ = ½ × 36 × π/3
✓ 1
A1 6π m2
✓ 1
Part (c)
A1 12 + 2π m (≈ 18.2832 m); ECF from (a)
✓ 1
Exam tip: Formulas for arc length and sector area are only valid when the angle is in radians; keep π in your answer when the question asks for it in terms of π.
40IBMA3S-3-17r24-033.5 The unit circle and the ambiguous casePaper 1Easy5 marks
No calculator
The angle θ satisfies \(\frac{\pi}{2}<\theta<\pi\) and \(\sin\theta=\frac{5}{13}\).
(a)
Find the exact value of cos θ.
(3)
(b)
Find the exact value of sin 2θ.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of sin2θ + cos2θ = 1, giving cos2θ = 144/169
✓ 1
A1 cos θ = ±12/13
✓ 1
A1 cos θ = −12/13 since θ is in the second quadrant (cosine is negative)
✓ 1
Part (b)
M1 use of sin 2θ = 2 sin θ cos θ = 2 × (5/13) × (−12/13) (ECF from (a))
✓ 1
A1 −120/169 (≈ -0.710059)
✓ 1
Common error: When you find cos θ from the identity, state the sign using the quadrant; leaving the answer as ±12/13 loses the final mark.
41IBMA3S-3-17r25-023.6 Trigonometric identitiesPaper 1Easy6 marks
No calculator
The angle x is acute and sin x = \(\frac{2}{3}\).
(a)
Find the exact value of cos x.
(2)
(b)
Find the exact value of sin 2x.
(2)
(c)
Find the exact value of cos 2x.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use sin2x + cos2x = 1 to get cos2x = 1 − \(\frac{4}{9}\) = \(\frac{5}{9}\)
✓ 1
A1 cos x = \(\frac{\sqrt{5}}{3}\) (positive because x is acute)
✓ 1
Part (b)
M1 use sin 2x = 2 sin x cos x
✓ 1
A1 sin 2x = \(\frac{4\sqrt{5}}{9}\) (ECF from (a))
✓ 1
Part (c)
M1 use cos 2x = 1 − 2sin2x = 1 − 2\(\left(\frac{2}{3}\right)^2\)
✓ 1
A1 cos 2x = \(\frac{1}{9}\)
✓ 1
Common error: Remember to check the sign when taking a square root: an acute angle has positive sine and cosine, so only the positive root is correct.
42IBMA3S-3-19r30-023.6 Trigonometric identitiesPaper 1Easy6 marks
No calculator
The angle θ is obtuse and sin θ = 2⁄3.
(a)
Find the exact value of cos θ.
(2)
(b)
Find the exact value of cos 2θ.
(2)
(c)
Find the exact value of sin 2θ.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of sin²θ + cos²θ = 1, giving cos²θ = 5⁄9
✓ 1
A1 cos θ = −√5⁄3 (negative because θ is obtuse)
✓ 1
Part (b)
M1 use of cos 2θ = 1 − 2sin²θ
✓ 1
A1 cos 2θ = 1⁄9
✓ 1
Part (c)
M1 use of sin 2θ = 2 sin θ cos θ
✓ 1
A1 sin 2θ = −4√5⁄9 (≈ −0.9938)
✓ 1
Common error: When an angle is given as obtuse, the cosine is negative; taking the positive square root loses the final mark.
43IBMA3S-3-19r31-023.4 Radian measure, arcs and sectorsPaper 1Easy6 marks
No calculator
A cooling tray in a coffee roaster is a sector of a circle with centre O and radius \(r\) metres. The angle of the sector is \(\frac{\pi}{3}\) radians and its area is \(0.24\pi\ \text{m}^2\).
Diagram not to scale
(a)
Find the value of \(r\).
(3)
(b)
Find the arc length of the sector, giving your answer in terms of \(\pi\).
(1)
(c)
Find the perimeter of the sector, giving your answer in the form \(a+b\pi\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of area \(=\frac{1}{2}r^2\theta\)
✓ 1
M1 \(\frac{1}{2}r^2\cdot\frac{\pi}{3}=0.24\pi\), so \(r^2=1.44\)
✓ 1
A1 \(r=1.2\)
✓ 1
Part (b)
A1 arc length \(=1.2\times\frac{\pi}{3}=0.4\pi\) m
✓ 1
Part (c)
M1 perimeter \(=2r+\) arc length
✓ 1
A1 \(2.4+0.4\pi\) m
✓ 1
Exam tip: The sector formulas arc \(=r\theta\) and area \(=\frac{1}{2}r^2\theta\) only work when the angle is in radians.
44IBMA3S-3-19r31-133.8 Trigonometric equationsPaper 1Easy6 marks
No calculator
The strength of a signal from a spinning probe is modelled by \(S(t)=4\cos 2t+1\), where \(t\) is in seconds and \(0\le t\le\pi\).
(a)
Write down the amplitude and the period of \(S\).
(2)
(b)
Solve \(S(t)=3\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 amplitude = 4
✓ 1
A1 period \(=\pi\) s
✓ 1
Part (b)
M1 \(\cos 2t=\frac{1}{2}\)
✓ 1
M1 range for \(2t\) is \(0\le 2t\le 2\pi\)
✓ 1
A1 \(2t=\frac{\pi}{3}\) or \(\frac{5\pi}{3}\)
✓ 1
A1 \(t=\frac{\pi}{6}\) or \(\frac{5\pi}{6}\)
✓ 1
Common error: When the argument is \(2t\), the interval for \(2t\) is twice as long as the interval for \(t\); forgetting this loses the second solution.
45IBMA3S-3-17r48-043.1 Three-dimensional geometryPaper 1Easy4 marks
No calculator
A wedge is cut from a circular cheese of radius 6 cm. The top face of the wedge is a sector of the circle with angle \(\frac{\pi}{3}\) radians at the centre.
Diagram not to scale
(a)
Find the length of the curved edge of the top face, giving your answer in terms of \(\pi\).
(2)
(b)
Find the area of the top face, giving your answer in terms of \(\pi\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use \(l = r\theta\) with \(r = 6\) and \(\theta = \frac{\pi}{3}\)
Common error: The formulas \(l = r\theta\) and \(A = \frac{1}{2}r^2\theta\) only work when \(\theta\) is in radians.
46AASL3-q00053.1 Three-dimensional geometryPaper 1Medium8 marks
A solid hemisphere and a solid right circular cone both have base radius \(r\) cm. The total surface area of the hemisphere (including its flat face) is equal to the total surface area of the cone (including its base).
(a)
Show that the slant height of the cone is \(2r\) cm.
(2)
(b)
Find the exact height of the cone in terms of \(r\).
(1)
(c)
Show that \(\dfrac{\text{volume of the cone}}{\text{volume of the hemisphere}}=\dfrac{\sqrt{3}}{2}\).
(3)
(d)
Find the exact angle, in radians, between a slant edge of the cone and its base.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(2\pi r^{2}+\pi r^{2}=\pi r^{2}+\pi rl\)
M1
Hemisphere: curved 2πr² plus flat πr²; cone: base plus curved surface.
\(3\pi r^{2}-\pi r^{2}=\pi rl\) or \(2\pi r^{2}=\pi rl\)
49AASL3-q00173.2 Right-angled and non-right-angled trigonometryPaper 1Medium7 marks
The lengths of the sides of a triangle are \(x\) cm, \((x+1)\) cm and \((x+2)\) cm, where \(x\gt0\). The largest angle of the triangle is \(120^{\circ}\).
Diagram not to scale
(a)
Show that \(2x^{2}-x-3=0\).
(3)
(b)
Find the value of \(x\).
(2)
(c)
Find the exact area of the triangle.
(2)
Show mark scheme
Working
Mark
Notes
(a)
the largest angle is opposite the longest side: \((x+2)^{2}=x^{2}+(x+1)^{2}-2x(x+1)\cos120^{\circ}\)
M1
\(\cos120^{\circ}=-\dfrac{1}{2}\), so \((x+2)^{2}=x^{2}+(x+1)^{2}+x(x+1)\)
50AASL3-q00243.3 Applications of trigonometryPaper 1Medium6 marks
A surveyor stands at point \(P\) on horizontal ground and measures the angle of elevation of the top of a vertical tower as \(45^{\circ}\). She then walks \(18\) m directly towards the base of the tower to point \(Q\), where the angle of elevation of the top of the tower is \(60^{\circ}\). The height of the tower is \(h\) m.
Diagram not to scale
(a)
Show that \(h=27+9\sqrt{3}\).
(4)
(b)
Find the exact distance from \(Q\) to the base of the tower.
(2)
Show mark scheme
Working
Mark
Notes
(a)
distance from \(P\) to the base \(=\dfrac{h}{\tan45^{\circ}}=h\)
M1
distance from \(Q\) to the base \(=\dfrac{h}{\tan60^{\circ}}=\dfrac{h}{\sqrt{3}}\)
51AASL3-q00363.4 Radian measure, arcs and sectorsPaper 1Medium7 marks
The diagram shows a sector \(OAB\) of a circle with centre \(O\) and radius \(r\) cm, where \(A\hat{O}B=\dfrac{\pi}{3}\). The shaded region is the segment bounded by the chord \([AB]\) and the arc \(AB\).
(a)
Show that the area of the shaded region is \(\dfrac{r^{2}}{12}\left(2\pi-3\sqrt{3}\right)\text{ cm}^{2}\).
(3)
(b)
The area of the shaded region is \(\left(6\pi-9\sqrt{3}\right)\text{ cm}^{2}\). Find the value of \(r\).
(2)
(c)
Hence find the exact perimeter of the shaded region.
(2)
Show mark scheme
Working
Mark
Notes
(a)
area of sector \(=\dfrac{1}{2}r^{2}\left(\dfrac{\pi}{3}\right)=\dfrac{\pi r^{2}}{6}\)
M1
area of triangle \(=\dfrac{1}{2}r^{2}\sin\dfrac{\pi}{3}=\dfrac{\sqrt{3}}{4}r^{2}\)
Award A0 if \(r=-6\) is also given as a final answer.
(c)
arc \(=6\times\dfrac{\pi}{3}=2\pi\) and chord \(AB=6\) (triangle \(OAB\) is equilateral)
(M1)
FT from (b).
\(\left(2\pi+6\right)\) cm
A1
Answers: (b) \(r=6\) · (c) \(2\pi+6\) cm
52AASL3-q00383.4 Radian measure, arcs and sectorsPaper 1Medium6 marks
A sector of a circle has radius \(15\) cm and angle \(\theta\) radians. The two straight edges of the sector are joined, without overlap, to form the curved surface of a right circular cone with base radius \(9\) cm.
Diagram not to scale
(a)
Find the exact value of \(\theta\).
(2)
(b)
Find the height of the cone.
(2)
(c)
Find the exact volume of the cone.
(2)
Show mark scheme
Working
Mark
Notes
(a)
arc length of the sector = circumference of the base: \(15\theta=2\pi(9)\)
(M1)
\(\theta=\dfrac{6\pi}{5}\)
A1
(b)
slant height \(=15\): \(h=\sqrt{15^{2}-9^{2}}\)
(M1)
Recognizing the radius of the sector becomes the slant height.
53AASL3-q00393.4 Radian measure, arcs and sectorsPaper 1Medium9 marks
Diagram not to scale
The diagram shows a circle with centre \(O\) and radius \(r\) cm. The point \(T\) is such that \(OT=2r\) cm. The lines \((TA)\) and \((TB)\) are tangents to the circle at \(A\) and \(B\). The shaded region is bounded by \([TA]\), \([TB]\) and the minor arc \(AB\).
(a)
Show that \(A\hat{O}T=\dfrac{\pi}{3}\).
(2)
(b)
Show that the area of the shaded region is \(\left(\sqrt{3}-\dfrac{\pi}{3}\right)r^{2}\text{ cm}^{2}\).
(3)
(c)
Find an expression, in terms of \(r\), for the exact perimeter of the shaded region.
(2)
(d)
For one value of \(r\), the area of the shaded region in cm2 is numerically equal to its perimeter in cm. Show that this value is \(r=\dfrac{2\left(3\sqrt{3}+\pi\right)}{3\sqrt{3}-\pi}\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(O\hat{A}T=\dfrac{\pi}{2}\) (tangent perpendicular to radius), so \(\cos A\hat{O}T=\dfrac{r}{2r}\)
M1
\(\cos A\hat{O}T=\dfrac{1}{2}\)
A1
\(A\hat{O}T=\dfrac{\pi}{3}\)
AG
(b)
area of kite \(OATB=2\times\dfrac{1}{2}(r)\left(\sqrt{3}\,r\right)=\sqrt{3}\,r^{2}\)
M1
\(AT=\sqrt{4r^{2}-r^{2}}=\sqrt{3}\,r\).
area of sector \(OAB=\dfrac{1}{2}r^{2}\left(\dfrac{2\pi}{3}\right)=\dfrac{\pi r^{2}}{3}\)
M1
\(\sqrt{3}\,r^{2}-\dfrac{\pi r^{2}}{3}\)
A1
Must see the subtraction.
\(\left(\sqrt{3}-\dfrac{\pi}{3}\right)r^{2}\)
AG
(c)
\(TA=TB=\sqrt{3}\,r\) and arc \(AB=r\left(\dfrac{2\pi}{3}\right)\)
(M1)
Award for either correct length.
\(\left(2\sqrt{3}+\dfrac{2\pi}{3}\right)r\) cm
A1
Accept equivalent forms, e.g. \(\dfrac{2r\left(3\sqrt{3}+\pi\right)}{3}\).
(d)
\(\left(\sqrt{3}-\dfrac{\pi}{3}\right)r^{2}=\left(2\sqrt{3}+\dfrac{2\pi}{3}\right)r\) and dividing by \(r\;(r\neq0)\)
Answers: (c) \(\left(2\sqrt{3}+\dfrac{2\pi}{3}\right)r\) cm
54AASL3-q00453.5 The unit circle and the ambiguous casePaper 1Medium7 marks
The point \(P\left(-\dfrac{5}{13},\,\dfrac{12}{13}\right)\) lies on the unit circle with centre \(O\). The angle between the positive \(x\)-axis and \([OP]\), measured anticlockwise, is \(\theta\).
(a)
Write down the value of \(\tan\theta\).
(1)
(b)
Find the value of \(\sin2\theta\).
(2)
(c)
Find the value of \(\cos2\theta\).
(2)
(d)
The point \(Q\) on the unit circle is such that the angle between the positive \(x\)-axis and \([OQ]\), measured anticlockwise, is \(\theta+\pi\). Write down the coordinates of \(Q\).
55AASL3-q00483.5 The unit circle and the ambiguous casePaper 1Medium6 marks
The line \(L\) passes through the origin \(O\) and makes an angle of \(\dfrac{2\pi}{3}\) with the positive \(x\)-axis, measured anticlockwise.
(a)
Find the equation of \(L\), giving your answer in the form \(y=mx\), where \(m\) is exact.
(2)
(b)
The line \(L\) meets the unit circle \(x^{2}+y^{2}=1\) at the points \(A\) and \(B\), where \(A\) lies in the second quadrant. Find the exact coordinates of \(A\) and of \(B\).
(2)
(c)
The line \(M\) has equation \(y=x\). Find the acute angle between \(L\) and \(M\).
57AASL3-q00503.5 The unit circle and the ambiguous casePaper 1Medium9 marks
The point \(P(\cos\theta,\,\sin\theta)\) lies on the unit circle with centre \(O\), where \(0\lt\theta\lt\dfrac{\pi}{2}\). The point \(Q\) is the reflection of \(P\) in the \(y\)-axis and the point \(R\) is the reflection of \(P\) in \(O\).
Diagram not to scale
(a)
Write down the coordinates of \(Q\) and of \(R\) in terms of \(\theta\).
(2)
(b)
Show that the area of triangle \(PQR\) is \(2\sin\theta\cos\theta\).
(2)
(c)
The points \(Q\) and \(R\) can also be written as \(Q(\cos\alpha,\,\sin\alpha)\) and \(R(\cos\beta,\,\sin\beta)\), where \(0\lt\alpha\lt2\pi\) and \(0\lt\beta\lt2\pi\). Write down \(\alpha\) and \(\beta\) in terms of \(\theta\).
(2)
(d)
Given that the gradient of the line \((RP)\) is \(\dfrac12\), find the exact area of triangle \(PQR\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(Q(-\cos\theta,\,\sin\theta)\)
A1
\(R(-\cos\theta,\,-\sin\theta)\)
A1
(b)
\(PQ=2\cos\theta\) and \(QR=2\sin\theta\), with a right angle at \(Q\)
M1
area \(=\dfrac12(2\cos\theta)(2\sin\theta)\)
A1
\(=2\sin\theta\cos\theta\)
AG
(c)
\(\alpha=\pi-\theta\)
A1
\(\beta=\pi+\theta\)
A1
(d)
\((RP)\) passes through \(O\), so \(\tan\theta=\dfrac12\)
(M1)
line through the origin: \(y=x\tan\theta\)
\(\sin\theta=\dfrac{1}{\sqrt5}\) and \(\cos\theta=\dfrac{2}{\sqrt5}\)
(M1)
e.g. from a right-angled triangle with sides 1, 2, \(\sqrt5\) (\(\theta\) acute).
area \(=2\left(\dfrac{1}{\sqrt5}\right)\left(\dfrac{2}{\sqrt5}\right)=\dfrac45\)
A1
Or \(\dfrac12(2\cos\theta)(2\sin\theta)\) with their values.
59AASL3-q00603.6 Trigonometric identitiesPaper 1Medium9 marks
The diagram shows triangle \(ABC\) with \(AB=AC=6\) cm and \(B\hat{A}C=2\theta\), where \(\cos\theta=\dfrac{2}{3}\) and \(0\lt\theta\lt\dfrac{\pi}{2}\).
Diagram not to scale
(a)
Show that \(\sin2\theta=\dfrac{4\sqrt5}{9}\).
(3)
(b)
Hence find the exact area of triangle \(ABC\).
(2)
(c)
Find the exact value of \(\cos2\theta\).
(2)
(d)
Hence, or otherwise, find the exact length of \([BC]\).
62AASL3-q00663.7 Circular functions and their graphsPaper 1Medium7 marks
Let \(f(x)=a\sin\big(b(x+c)\big)+d\), where \(a,b\gt0\) and \(0\lt c\lt\dfrac{\pi}{2}\). The diagram shows the graph of \(y=f(x)\) for \(0\le x\le\pi\). The graph has a maximum point at \(\left(\dfrac{\pi}{6},\,5\right)\) and a minimum point at \(\left(\dfrac{2\pi}{3},\,-1\right)\).
(a)
Write down the value of \(a\) and the value of \(d\).
(2)
(b)
Find the value of \(b\).
(2)
(c)
Find the value of \(c\).
(2)
(d)
Write down the number of solutions of the equation \(f(x)=4\) for \(0\le x\le\pi\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(a=3\)
A1
\(d=2\)
A1
(b)
period \(=2\left(\dfrac{2\pi}{3}-\dfrac{\pi}{6}\right)=\pi\)
(M1)
\(b=\dfrac{2\pi}{\pi}=2\)
A1
(c)
\(2\left(\dfrac{\pi}{6}+c\right)=\dfrac{\pi}{2}\)
(M1)
Or substituting \(\left(\frac{\pi}{6},5\right)\): \(3\sin\big(2(\frac{\pi}{6}+c)\big)+2=5\). FT from (a), (b).
64AASL3-q00703.7 Circular functions and their graphsPaper 1Medium9 marks
Let \(f(x)=2\cos2x+1\), for \(0\le x\le\pi\).
(a)
Find the \(x\)-intercepts of the graph of \(f\).
(3)
(b)
Sketch the graph of \(y=f(x)\), showing the coordinates of the end points and of the minimum point.
(3)
(c)
Hence find the set of values of \(x\) for which \(f(x)\gt2\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\cos2x=-\dfrac12\), with \(0\le2x\le2\pi\)
M1
\(2x=\dfrac{2\pi}{3},\ \dfrac{4\pi}{3}\)
(A1)
\(x=\dfrac{\pi}{3},\ \dfrac{2\pi}{3}\)
A1
(b)
end points \((0,\,3)\) and \((\pi,\,3)\)
A1
minimum point \(\left(\dfrac{\pi}{2},\,-1\right)\)
A1
correct shape: smooth curve symmetric about \(x=\dfrac{\pi}{2}\), crossing the \(x\)-axis at their intercepts
A1
(c)
\(\cos2x=\dfrac12\)
M1
\(x=\dfrac{\pi}{6},\ \dfrac{5\pi}{6}\)
A1
\(0\le x\lt\dfrac{\pi}{6}\) or \(\dfrac{5\pi}{6}\lt x\le\pi\)
A1
Accept strict or non-strict inequality at 0 and π; endpoints π/6 and 5π/6 must be excluded.
Answers: (a) \(\frac{\pi}{3},\frac{2\pi}{3}\) · (c) \(0\le x\lt\frac{\pi}{6}\) or \(\frac{5\pi}{6}\lt x\le\pi\)
Model sketch
65AASL3-q00713.7 Circular functions and their graphsPaper 1Medium10 marks
The graph of \(y=\sin x\) is transformed onto the graph of \(y=g(x)\) by the following sequence of transformations:
a vertical stretch with scale factor \(3\);
a horizontal stretch with scale factor \(\dfrac12\);
a translation by the vector \(\begin{pmatrix}\frac{\pi}{6}\\-1\end{pmatrix}\).
(a)
Find \(g(x)\), giving your answer in the form \(g(x)=a\sin\big(b(x+c)\big)+d\).
(3)
(b)
The point \(\left(\dfrac{\pi}{2},\,1\right)\) on the graph of \(y=\sin x\) is mapped to the point \(M\) on the graph of \(y=g(x)\). Find the coordinates of \(M\).
(2)
(c)
Find the exact value of \(g(0)\).
(2)
(d)
Solve the equation \(g(x)=\dfrac12\), for \(0\le x\le\pi\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(a=3\)
A1
\(b=2\)
A1
\(c=-\dfrac{\pi}{6}\) and \(d=-1\), so \(g(x)=3\sin\left(2\left(x-\dfrac{\pi}{6}\right)\right)-1\)
A1
Award A0 for \(3\sin\left(2x-\frac{\pi}{6}\right)-1\).
Exam tip: Double angle identities apply to any angle: \(4\theta\) is double \(2\theta\), so \(\sin 4\theta=2\sin 2\theta\cos 2\theta\).
69IBMA2S-048-05-063.8 Trigonometric equationsPaper 1Medium6 marks
No calculator
Two models for the displacement of a swinging gate are equal when \(\cos 2x=3\sin x-1\). Solve this equation for \(0\le x\le 2\pi\), giving exact answers.
(a)
Solve \(\cos 2x=3\sin x-1\) for \(0\le x\le 2\pi\).
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitutes \(\cos 2x=1-2\sin^2x\)
✓ 1
A1 \(2\sin^2x+3\sin x-2=0\)
✓ 1
M1 factorises or uses formula: \((2\sin x-1)(\sin x+2)=0\)
✓ 1
A1 \(\sin x=\frac12\)
✓ 1
R1 rejects \(\sin x=-2\) since \(-1\le\sin x\le1\)
✓ 1
A1 \(x=\frac{\pi}{6},\ \frac{5\pi}{6}\)
✓ 1
Exam tip: When an equation mixes \(\cos 2x\) and \(\sin x\), choose \(\cos 2x=1-2\sin^2x\) so that only one trigonometric function remains.
70IBMA2S-048-05-083.6 Trigonometric identitiesPaper 1Medium6 marks
No calculator
Let \(g(x)=\cos^4x-\sin^4x\).
(a)
Show that \(g(x)=\cos 2x\).
(3)
(b)
Hence solve \(g(x)=-\frac12\) for \(0\le x\le\pi\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 difference of two squares: \((\cos^2x-\sin^2x)(\cos^2x+\sin^2x)\)
✓ 1
A1 \(\cos^2x+\sin^2x=1\)
✓ 1
A1 \(\cos^2x-\sin^2x=\cos 2x\)
✓ 1
Part (b)
M1 \(\cos 2x=-\frac12\) with \(0\le 2x\le 2\pi\)
✓ 1
A1 \(2x=\frac{2\pi}{3},\ \frac{4\pi}{3}\)
✓ 1
A1 \(x=\frac{\pi}{3},\ \frac{2\pi}{3}\)
✓ 1
Exam tip: Spot a difference of two squares in fourth powers: \(a^4-b^4=(a^2-b^2)(a^2+b^2)\) often unlocks the Pythagorean identity.
71IBMA2S-048-05-103.6 Trigonometric identitiesPaper 1Medium7 marks
No calculator
A tracking telescope rotates through an angle \(\theta\), where \(\frac{3\pi}{2}<\theta<2\pi\) and \(\tan\theta=-\frac{\sqrt5}{2}\).
(a)
Find the exact value of \(\cos\theta\) and of \(\sin\theta\).
(3)
(b)
Find the exact value of \(\cos 2\theta\).
(2)
(c)
Find the exact value of \(\sin 2\theta\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(\tan\theta=\frac{\sin\theta}{\cos\theta}\) with \(\sin^2\theta+\cos^2\theta=1\) (or a right triangle with hypotenuse 3)
✓ 1
A1 \(\cos\theta=\frac23\), positive in the fourth quadrant
Exam tip: In the fourth quadrant cosine is positive and sine is negative. Fix both signs before using any double angle formula.
72IBMA2S-048-07-053.1 Three-dimensional geometryPaper 1Medium7 marks
No calculator
A horizontal drainage pipe has a circular cross-section of radius 6 cm. Water lies in the bottom of the pipe to a depth of 3 cm. The water surface forms a chord AB of the circular cross-section, and O is the centre of the circle.
Cross-section, not to scale
(a)
Show that angle AOB = \(\frac{2\pi}{3}\).
(2)
(b)
Find the exact area of the cross-section occupied by the water.
(3)
(c)
Find the exact width of the water surface.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 perpendicular distance from O to AB is 6 − 3 = 3, so cos(½AÔB) = 3/6
✓ 1
A1 ½AÔB = π/3, so AÔB = 2π/3
✓ 1
Part (b)
M1 sector area ½ × 36 × 2π/3 = 12π
✓ 1
M1 subtracts triangle area ½ × 36 × sin(2π/3) = 9√3
✓ 1
A1 12π − 9√3 cm²
✓ 1
Part (c)
M1 AB = 2 × 6 sin(π/3), or uses the cosine rule AB² = 36 + 36 − 72cos(2π/3)
✓ 1
A1 6√3 cm
✓ 1
Exam tip: Segment area = ½r²(θ − sin θ). Learn the exact values sin(2π/3) = √3/2 and cos(2π/3) = −½, because you will need them without a calculator.
73IBMA2S-048-07-073.4 Radian measure, arcs and sectorsPaper 1Medium6 marks
No calculator
A disposable cone-shaped paper cup is made by joining the two straight edges of a paper sector of radius 12 cm and central angle θ radians, with no overlap. The finished cup is a right circular cone with base radius 4 cm.
Paper sector, not to scale
(a)
Find θ.
(2)
(b)
Find the exact height of the cup.
(2)
(c)
Find the exact volume of the cup.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 arc length of sector equals the base circumference: 12θ = 2π × 4
✓ 1
A1 θ = 2π/3
✓ 1
Part (b)
M1 slant height is 12, so h = √(12² − 4²)
✓ 1
A1 h = √128 = 8√2 cm
✓ 1
Part (c)
M1 V = ⅓π × 4² × 8√2
✓ 1
A1 128√2π/3 cm³
✓ 1
Common error: The radius of the sector becomes the slant height of the cone, not its vertical height. The sector's arc becomes the circumference of the base.
74IBMA2S-048-02-063.6 Trigonometric identitiesPaper 1Medium7 marks
No calculator
A wheelchair ramp is inclined at an acute angle α to the horizontal, where sin α = k and 0 < k < 1. Give each answer in terms of k.
(a)
Find cos α.
(2)
(b)
Write down sin(π − α).
(1)
(c)
Find cos(π + α).
(2)
(d)
Find tan(−α).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses sin2α + cos2α = 1 to get cos2α = 1 − k2
✓ 1
A1 cos α = √(1 − k2), taking the positive root since α is acute
✓ 1
Part (b)
A1 sin(π − α) = k
✓ 1
Part (c)
M1 uses cos(π + α) = −cos α. Accept the quoted identity, the expansion cos π cos α − sin π sin α, or unit-circle symmetry
✓ 1
A1 cos(π + α) = −√(1 − k2)
✓ 1
Part (d)
M1 uses tan(−α) = −tan α = −sin α / cos α, or sin(−α)/cos(−α) = −k / cos α
✓ 1
A1 tan(−α) = −k / √(1 − k2)
✓ 1
Exam tip: Sketch the unit circle and mark the points at α, π − α, π + α and −α. The symmetry shows straight away which coordinates change sign.
75IBMA2S-048-02-073.5 The unit circle and the ambiguous casePaper 1Medium7 marks
No calculator
A triangular support frame ABC for a solar panel has AB = 6 m, AC = 3√2 m and angle ABC = π/6 radians. All angles are to be given in radians.
(a)
Find the possible values of angle ACB, justifying that each is valid.
(5)
(b)
Hence find the possible values of angle BAC.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sine rule: sin C / 6 = sin(π/6) / (3√2)
✓ 1
A1 sin C = 3/(3√2) = √2/2
✓ 1
A1 C = π/4
✓ 1
A1 C = 3π/4
✓ 1
R1 both valid since π/6 + 3π/4 = 11π/12 < π
✓ 1
Part (b)
A1 7π/12 (when C = π/4)
✓ 1
A1 π/12 (when C = 3π/4)
✓ 1
Common error: Stopping at C = π/4 loses marks: since AC < AB, the side opposite the known angle is the shorter one, so check whether the obtuse angle π − C also fits.
76IBMA2S-048-03-033.8 Trigonometric equationsPaper 1Medium6 marks
No calculator
The curves \(y = 2\cos^2 x\) and \(y = 3 - 3\sin x\) are defined for \(0 \le x \le 2\pi\). Find the exact x-coordinates of all points where the two curves meet.
(a)
Find the exact x-coordinates of the points of intersection.
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 equates the curves and uses \(\cos^2 x = 1 - \sin^2 x\)
✓ 1
A1 \(2\sin^2 x - 3\sin x + 1 = 0\)
✓ 1
M1 factorises: \((2\sin x - 1)(\sin x - 1) = 0\)
✓ 1
A1 \(\sin x = \frac{1}{2}\) or \(\sin x = 1\)
✓ 1
A1 \(x = \frac{\pi}{2}\)
✓ 1
A1 \(x = \frac{\pi}{6}, \frac{5\pi}{6}\)
✓ 1
Exam tip: When an equation mixes \(\sin x\) and \(\cos^2 x\), replace \(\cos^2 x\) by \(1 - \sin^2 x\). This gives a quadratic in a single trigonometric ratio.
77IBMA2S-048-03-043.8 Trigonometric equationsPaper 1Medium6 marks
No calculator
In an electronics test, two voltage signals are modelled by \(V_1 = \sin 2t\) and \(V_2 = \sqrt{3}\sin t\), for \(0 \le t \le 2\pi\).
(a)
Show that the signals are equal when \(\sin t\,(2\cos t - \sqrt{3}) = 0\).
(2)
(b)
Hence find all values of t for which \(V_1 = V_2\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(\sin 2t = 2\sin t\cos t\)
✓ 1
A1 \(2\sin t\cos t - \sqrt{3}\sin t = 0 \Rightarrow \sin t\,(2\cos t - \sqrt{3}) = 0\)
✓ 1
Part (b)
M1 sets each factor equal to zero: \(\sin t = 0\) or \(\cos t = \frac{\sqrt{3}}{2}\)
✓ 1
A1 \(t = 0, \pi, 2\pi\)
✓ 1
A1 \(t = \frac{\pi}{6}\)
✓ 1
A1 \(t = \frac{11\pi}{6}\)
✓ 1
Common error: Do not divide both sides by \(\sin t\). Doing so loses the solutions where \(\sin t = 0\). Factorise instead, so that both factors give solutions.
78IBMA2S-048-03-083.8 Trigonometric equationsPaper 1Medium7 marks
No calculator
Let \(f(x) = 3\sin 2x + 1\), for \(0 \le x \le \pi\).
(a)
Find the range of f.
(2)
(b)
Solve \(f(x) = -\frac{1}{2}\).
(3)
(c)
Find the set of values of k for which the equation \(f(x) = k\) has exactly two solutions.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 minimum −2
✓ 1
A1 range \(-2 \le f(x) \le 4\)
✓ 1
Part (b)
M1 \(\sin 2x = -\frac{1}{2}\)
✓ 1
A1 \(2x = \frac{7\pi}{6}, \frac{11\pi}{6}\)
✓ 1
A1 \(x = \frac{7\pi}{12}, \frac{11\pi}{12}\)
✓ 1
Part (c)
R1 recognises that \(k = 1\) gives three solutions (\(x = 0, \frac{\pi}{2}, \pi\)) and that \(k = -2\) or \(k = 4\) gives one solution
✓ 1
A1 \(-2 < k < 1\) or \(1 < k < 4\)
✓ 1
Common error: Remember the endpoints of the domain. At \(k = 1\) the line meets the graph at \(x = 0\), at \(x = \frac{\pi}{2}\) and at \(x = \pi\), so this value must be excluded.
79IBMA2S-048-00-053.2 Right-angled and non-right-angled trigonometryPaper 1Medium6 marks
No calculator
A sports hall is modelled as a cuboid ABCDEFGH. The rectangular floor ABCD has AB = 12 m and BC = 4 m. The vertical edges AE, BF, CG and DH each have length 3 m. A laser beam is fired from floor corner A to ceiling corner G.
Diagram not to scale
(a)
Find the length of the beam AG.
(2)
(b)
Find the exact value of the tangent of the angle between the beam and the floor.
(2)
(c)
Find the exact value of the cosine of the angle between the beam AG and the floor edge AB.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 AC2 = 122 + 42 = 160, then AG2 = 160 + 32
✓ 1
A1 AG = 13 m
✓ 1
Part (b)
M1 identifies angle GAC, with tan = CG/AC = 3/√160
✓ 1
A1 3/(4√10) = 3√10/40
✓ 1
Part (c)
R1 triangle ABG is right-angled at B, since AB is perpendicular to the wall BCGF
✓ 1
A1 cos GAB = AB/AG = 12/13
✓ 1
Exam tip: The angle between a line and a plane is measured to the line's projection on the plane. Here the projection of AG on the floor is AC, not AB.
80IBMA2S-048-00-083.1 Three-dimensional geometryPaper 1Medium6 marks
No calculator
A cable-suspended camera in a stadium moves in a straight line from A(−10, 4, 18) to B. Coordinates are in metres, the pitch lies in the plane z = 0 and the z-axis is vertical. The midpoint of the camera's path is M(2, 0, 21).
(a)
Find the coordinates of B.
(2)
(b)
Find the length of the camera's path AB.
(2)
(c)
Find the exact value of the sine of the angle between the path AB and the pitch.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses midpoint, e.g. (−10 + x)/2 = 2
✓ 1
A1 B(14, −4, 24)
✓ 1
Part (b)
M1 √(242 + (−8)2 + 62) = √676
✓ 1
A1 26 m
✓ 1
Part (c)
M1 vertical rise 24 − 18 = 6 m opposite the angle, hypotenuse 26 m
✓ 1
A1 sin θ = 6/26 = 3/13
✓ 1
Common error: To find B from a midpoint, double the midpoint and subtract A: B = 2M − A. Averaging A and M gives a quarter-way point instead.
81IBMA2S-048-01-043.8 Trigonometric equationsPaper 1Medium7 marks
No calculator
Let \(f(x)=5-2\sin\left(\frac{x}{2}\right)\) for \(0\le x\le 4\pi\).
(a)
Write down the period of \(f\).
(1)
(b)
Find the coordinates of the minimum point of the graph of \(f\).
(3)
(c)
Find the set of values of \(p\) for which the equation \(f(x)=p\) has exactly two solutions.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(4\pi\)
✓ 1
Part (b)
M1 minimum when \(\sin\left(\frac{x}{2}\right)=1\)
✓ 1
A1 \(x=\pi\)
✓ 1
A1 \(y=3\), so \((\pi,\,3)\)
✓ 1
Part (c)
R1 \(f(0)=f(2\pi)=f(4\pi)=5\) so \(p=5\) gives three solutions, while \(p=3\) or \(p=7\) gives one solution
✓ 1
A1 \(3<p<7\)
✓ 1
A1 \(p\ne 5\)
✓ 1
Exam tip: Sketch the graph across the whole domain and include its endpoints: horizontal lines through an endpoint value often change the number of intersections.
82IBMA2S-048-01-083.8 Trigonometric equationsPaper 1Medium7 marks
No calculator
Let \(f(x)=4\sin\left(x+\frac{\pi}{6}\right)\) for \(0\le x\le 2\pi\).
(a)
Write down the maximum value of \(f\) and the value of \(x\) at which it occurs.
(2)
(b)
Solve \(f(x)=2\).
(4)
(c)
Hence find the total length of the interval of \(x\) for which \(f(x)>2\).
A1 \(f(x)>2\) for \(0<x<\frac{2\pi}{3}\), length \(\frac{2\pi}{3}\)
✓ 1
Exam tip: Shift the interval for the whole argument (x + π/6) before solving. Both endpoints of the domain turn out to be solutions here, and they are easy to miss.
83IBMA2S-048-06-073.2 Right-angled and non-right-angled trigonometryPaper 1Medium7 marks
No calculator
A triangular bracket ABC supports a small solar panel. AB = 5 cm, AC = 3 cm and angle BAC = 120°.
A1 use of \(\cos 120° = -\frac{1}{2}\) giving \(BC^2 = 49\)
✓ 1
A1 BC = 7 cm
✓ 1
Part (b)
M1 \(\frac{1}{2}(5)(3)\sin 120°\) with \(\sin 120° = \frac{\sqrt{3}}{2}\)
✓ 1
A1 area = \(\frac{15\sqrt{3}}{4}\) cm2
✓ 1
Part (c)
M1 sine rule: \(\frac{\sin B}{3} = \frac{\sin 120°}{7}\)
✓ 1
A1 \(\sin A\hat{B}C = \frac{3\sqrt{3}}{14}\)
✓ 1
Exam tip: For angles in the second quadrant use \(\sin(180° - \theta) = \sin\theta\) and \(\cos(180° - \theta) = -\cos\theta\), so \(\sin 120° = \frac{\sqrt{3}}{2}\) and \(\cos 120° = -\frac{1}{2}\).
84IBMA2S-048-04-073.2 Right-angled and non-right-angled trigonometryPaper 1Medium7 marks
No calculator
Two straight struts in a roof truss are joined at one end, and the angle between them is 120°. One strut is 4 m longer than the other. A horizontal tie beam of length 14 m joins the free ends of the two struts, forming a triangle. Let the shorter strut have length x m.
Diagram not to scale
(a)
Show that \(x^2 + 4x - 60 = 0\).
(3)
(b)
Find the lengths of the two struts.
(2)
(c)
Find the exact area of the triangle formed by the struts and the tie beam.
A1 cancel \(2\sin\theta\) to obtain \(\frac{\sin\theta}{\cos\theta}=\tan\theta\) (AG)
✓ 1
Part (b)
M1 \(\tan\theta=\frac1{\sqrt3}\)
✓ 1
A1 \(\theta=\frac\pi6\) (≈ 0.524)
✓ 1
Exam tip: In a 'show that' question every step must appear; write the identity you use at each line.
87IBMA3S-3-17r8-073.8 Trigonometric equationsPaper 1Medium7 marks
No calculator
Tremor monitors on a volcano record an angle \(\theta\) (in degrees) that satisfies \(2\cos^2\theta+3\sin\theta=3\) for \(0^\circ\le\theta\le360^\circ\).
(a)
Show that the equation can be written as \(2\sin^2\theta-3\sin\theta+1=0\).
(3)
(b)
Hence find all possible values of \(\theta\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use \(\cos^2\theta=1-\sin^2\theta\)
✓ 1
M1 expand \(2-2\sin^2\theta+3\sin\theta=3\)
✓ 1
A1 rearrange to \(2\sin^2\theta-3\sin\theta+1=0\) (AG)
✓ 1
Part (b)
M1 factorise \((2\sin\theta-1)(\sin\theta-1)=0\)
✓ 1
A1 \(\sin\theta=\frac12\) or \(\sin\theta=1\)
✓ 1
A1 \(\theta=30^\circ,\ 150^\circ\)
✓ 1
A1 \(\theta=90^\circ\)
✓ 1
Exam tip: When both \(\sin\) and \(\cos^2\) appear, replace \(\cos^2\theta\) using the Pythagorean identity so the equation is a quadratic in one function.
88IBMA3S-3-16r18-063.4 Radian measure, arcs and sectorsPaper 1Medium6 marks
No calculator
A shallow area in a swimming pool is in the shape of a sector of a circle with radius 6 m. Its area is \(12\pi\) m2. The angle of the sector is \(\theta\) radians.
Diagram not to scale
(a)
Find the value of \(\theta\).
(3)
(b)
Find the perimeter of the sector, giving your answer in terms of \(\pi\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{1}{2}\times6^2\times\theta=12\pi\)
✓ 1
M1 \(18\theta=12\pi\)
✓ 1
A1 \(\theta=\frac{2\pi}{3}\)
✓ 1
Part (b)
M1 arc length \(=6\times\frac{2\pi}{3}\)
✓ 1
A1 arc length \(=4\pi\)
✓ 1
A1 perimeter \(=12+4\pi\) m (ECF from (a))
✓ 1
Common error: The perimeter of a sector includes the two radii as well as the arc.
89IBMA3S-3-16r18-083.8 Trigonometric equationsPaper 1Medium6 marks
No calculator
Solve the equation \(2\cos^2x+3\sin x=3\) for \(0\le x\le2\pi\).
(a)
Solve the equation, giving your answers in exact form.
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 replace \(\cos^2x\) with \(1-\sin^2x\)
✓ 1
A1 \(2s^2-3s+1=0\) where \(s=\sin x\)
✓ 1
M1 factorise or otherwise solve: \((2s-1)(s-1)=0\)
✓ 1
A1 \(\sin x=\frac{1}{2}\) or \(\sin x=1\)
✓ 1
A1 \(x=\frac{\pi}{6},\frac{5\pi}{6}\)
✓ 1
A1 \(x=\frac{\pi}{2}\)
✓ 1
Common error: Do not forget the second solution of \(\sin x=\frac{1}{2}\) in the interval, which is \(\pi-\frac{\pi}{6}\).
90IBMA3S-3-17r24-103.8 Trigonometric equationsPaper 1Medium7 marks
No calculator
Consider the equation \(2\sin^2 x = 1 + \cos x\), for \(0 \le x \le 2\pi\).
(a)
Show that the equation can be written as \(2\cos^2 x + \cos x - 1 = 0\).
(2)
(b)
Hence solve the equation, giving your solutions in terms of π.
(4)
(c)
Find the sum of all the solutions in (b).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 substitute sin2x = 1 − cos2x
✓ 1
A1 2 − 2cos2x = 1 + cos x, which rearranges to 2cos2x + cos x − 1 = 0 (AG)
✓ 1
Part (b)
M1 factorise or solve the quadratic in cos x: (2cos x − 1)(cos x + 1) = 0
✓ 1
A1 cos x = 1/2 or cos x = −1
✓ 1
A1 x = π/3 and 5π/3
✓ 1
A1 x = π
✓ 1
Part (c)
A1 π/3 + 5π/3 + π = 3π (≈ 9.424778); ECF from (b)
✓ 1
Exam tip: After solving for cos x, list every angle in the stated interval; cos x = 1/2 has two solutions in [0, 2π], and cos x = −1 has one.
91IBMA3S-3-17r25-073.6 Trigonometric identitiesPaper 1Medium7 marks
No calculator
Consider the expression \(\frac{1-\cos 2\theta}{\sin 2\theta}\), where sin 2θ ≠ 0.
(a)
Show that \(\frac{1-\cos 2\theta}{\sin 2\theta} \equiv \tan\theta\).
A1 cancel 2 sin θ and use tan θ = \(\frac{\sin\theta}{\cos\theta}\) to reach tan θ (AG)
✓ 1
Part (b)
M1 tan θ = √3
✓ 1
A1 θ = \(\frac{\pi}{3}\)
✓ 1
R1 no other solution in the interval, since tan has period π and \(\frac{\pi}{3}+\pi\) is outside it (tan θ is negative for \(\frac{\pi}{2}\) < θ < π)
✓ 1
Exam tip: In a 'show that' question, every line must be written down: state each identity you use before substituting it.
92IBMA3S-3-17r25-083.8 Trigonometric equationsPaper 1Medium6 marks
No calculator
Consider the equation 2cos2x + 3 sin x − 3 = 0, for 0 ≤ x ≤ 2π.
(a)
Show that the equation can be written as 2sin2x − 3 sin x + 1 = 0.
(2)
(b)
Hence solve the equation for 0 ≤ x ≤ 2π, giving exact values.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 replace cos2x by 1 − sin2x
✓ 1
A1 2 − 2sin2x + 3 sin x − 3 = 0, which simplifies to 2sin2x − 3 sin x + 1 = 0 (AG)
✓ 1
Part (b)
M1 attempt to factorise: (2 sin x − 1)(sin x − 1) = 0
✓ 1
A1 sin x = \(\frac{1}{2}\) or sin x = 1
✓ 1
A1 x = \(\frac{\pi}{6}\), \(\frac{5\pi}{6}\)
✓ 1
A1 x = \(\frac{\pi}{2}\)
✓ 1
Common error: When an equation contains both sin² and cos², rewrite it in a single function first; mixing the two cannot be factorised.
93IBMA3S-3-17r25-123.8 Trigonometric equationsPaper 1Medium9 marks
No calculator
A hospital monitor models a signal by f(x) = 3cos 2x + 1, for 0 ≤ x ≤ 2π, where x is in radians.
(a)
Write down the range of f.
(2)
(b)
Write down the period of f.
(1)
(c)
Solve f(x) = 2.5, giving exact values.
(4)
(d)
The equation f(x) = k has exactly two solutions for 0 ≤ x ≤ 2π. Find the value of k.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 amplitude 3 and vertical shift 1 used, e.g. maximum 3 + 1
✓ 1
A1 −2 ≤ f(x) ≤ 4
✓ 1
Part (b)
A1 period = π
✓ 1
Part (c)
M1 rearrange to cos 2x = \(\frac{1}{2}\)
✓ 1
A1 2x = \(\frac{\pi}{3}\)
✓ 1
M1 find further solutions for 0 ≤ 2x ≤ 4π, e.g. 2π − \(\frac{\pi}{3}\) and 2π + \(\frac{\pi}{3}\)
✓ 1
A1 x = \(\frac{\pi}{6}\), \(\frac{5\pi}{6}\), \(\frac{7\pi}{6}\), \(\frac{11\pi}{6}\)
✓ 1
Part (d)
M1 two solutions only occurs at the minimum: cos 2x = −1 gives x = \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\)
✓ 1
A1 k = −2
✓ 1
Exam tip: Replacing x by 2x squeezes the graph horizontally: you must find solutions for 2x over a doubled interval, then halve them.
94IBMA3S-3-17r25-133.8 Trigonometric equationsPaper 1Medium5 marks
No calculator
Solve 3 tan x = 2 sin x for 0 ≤ x ≤ 2π.
(a)
Solve the equation, giving exact answers.
(4)
(b)
A student divides both sides by sin x and concludes there are no solutions. Explain the error.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 replace tan x by \(\frac{\sin x}{\cos x}\) and multiply by cos x (cos x ≠ 0): 3 sin x = 2 sin x cos x
✓ 1
M1 factorise: sin x(3 − 2cos x) = 0
✓ 1
A1 sin x = 0 gives x = 0, π, 2π
✓ 1
R1 cos x = \(\frac{3}{2}\) has no solution, since |cos x| ≤ 1
✓ 1
Part (b)
R1 dividing by sin x removes the solutions where sin x = 0 (x = 0, π, 2π)
✓ 1
Common error: Never divide both sides of an equation by a trigonometric expression: you lose every solution where that expression is zero. Factorise instead.
95IBMA3S-3-19r27-063.8 Trigonometric equationsPaper 1Medium6 marks
No calculator
This question is about the identity \(\dfrac{1-\cos2x}{\sin2x}\equiv\tan x\).
(a)
Show that \(\dfrac{1-\cos2x}{\sin2x}\equiv\tan x\).
(3)
(b)
Hence solve \(\dfrac{1-\cos2x}{\sin2x}=\sqrt3\) for \(0<x<\pi\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(\cos2x=1-2\sin^2x\) to write the numerator as \(2\sin^2x\)
✓ 1
M1 use of \(\sin2x=2\sin x\cos x\) in the denominator
M1 sector area \(\frac12\times36\times\frac{\pi}{3}=6\pi\), minus triangle area
✓ 1
A1 \(6\pi-9\sqrt3\) cm2
✓ 1
Common error: The segment is the sector minus the triangle OAB, not minus a right-angled triangle; the triangle area needs \(\frac12r^2\sin\theta\).
97IBMA3S-3-19r27-103.8 Trigonometric equationsPaper 1Medium6 marks
No calculator
Solve \(2\cos^2x+3\sin x-3=0\) for \(0\le x\le2\pi\).
(a)
Solve \(2\cos^2x+3\sin x-3=0\) for \(0\le x\le2\pi\), giving exact answers.
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(\cos^2x=1-\sin^2x\)
✓ 1
A1 \(2\sin^2x-3\sin x+1=0\)
✓ 1
M1 attempt to factorise or solve, \((2\sin x-1)(\sin x-1)=0\)
✓ 1
A1 \(\sin x=\frac12\) or \(\sin x=1\)
✓ 1
A1 \(x=\frac{\pi}{6}\) and \(x=\frac{5\pi}{6}\)
✓ 1
A1 \(x=\frac{\pi}{2}\)
✓ 1
Exam tip: When an equation mixes \(\sin x\) and \(\cos^2x\), rewrite it using \(\sin^2x+\cos^2x=1\) so it becomes a quadratic in one function.
98IBMA3S-3-19r30-073.6 Trigonometric identitiesPaper 1Medium6 marks
No calculator
(a)
Show that (1 − cos 2θ) ⁄ sin 2θ ≡ tan θ.
(3)
(b)
Hence find the exact value of tan(π⁄8), giving your answer in the form √a − b.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of 1 − cos 2θ = 2sin²θ
✓ 1
M1 use of sin 2θ = 2 sin θ cos θ
✓ 1
A1 2sin²θ ⁄ (2 sin θ cos θ) = sin θ ⁄ cos θ = tan θ (AG)
✓ 1
Part (b)
M1 take θ = π⁄8 so that 2θ = π⁄4 and substitute cos(π⁄4) = sin(π⁄4) = √2⁄2
✓ 1
M1 (1 − √2⁄2) ⁄ (√2⁄2) = (2 − √2) ⁄ √2, then rationalise or simplify
✓ 1
A1 tan(π⁄8) = √2 − 1 (≈ 0.4142)
✓ 1
Exam tip: In a 'hence' question, look for the angle that turns the result you proved into the exact value you need.
99IBMA3S-3-19r30-103.8 Trigonometric equationsPaper 1Medium6 marks
No calculator
A model for interference in a telecoms cable leads to the equation 2sin²x + cos x = 1 for −π ≤ x ≤ π. Solve the equation, giving exact values of x.
(a)
Solve 2sin²x + cos x = 1 for −π ≤ x ≤ π.
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 replace sin²x by 1 − cos²x
✓ 1
M1 2cos²x − cos x − 1 = 0
✓ 1
M1 factorise or solve: (2cos x + 1)(cos x − 1) = 0
✓ 1
A1 cos x = 1 or cos x = −½
✓ 1
A1 x = 0
✓ 1
A1 x = ±2π⁄3
✓ 1
Exam tip: If an equation mixes sin² and cos, use sin²x + cos²x = 1 to write it in one function only.
100IBMA3S-3-19r31-073.6 Trigonometric identitiesPaper 1Medium7 marks
No calculator
The angle \(x\) through which a probe's solar panel is rotated satisfies \(\cos x=-\frac{2}{3}\), where \(\frac{\pi}{2}<x<\pi\).
(a)
Find the exact value of \(\sin x\).
(3)
(b)
Find the exact value of \(\sin 2x\).
(2)
(c)
Hence find the exact value of \(\tan 2x\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(\sin^2 x+\cos^2 x=1\), so \(\sin^2 x=\frac{5}{9}\)
✓ 1
R1 \(\sin x>0\) because \(x\) is in the second quadrant
✓ 1
A1 \(\sin x=\frac{\sqrt{5}}{3}\)
✓ 1
Part (b)
M1 use of \(\sin 2x=2\sin x\cos x\)
✓ 1
A1 \(\sin 2x=-\frac{4\sqrt{5}}{9}\)
✓ 1
Part (c)
M1 \(\cos 2x=2\cos^2x-1=-\frac{1}{9}\)
✓ 1
A1 \(\tan 2x=\frac{\sin 2x}{\cos 2x}=4\sqrt{5}\)
✓ 1
Common error: Taking \(\sin x=-\frac{\sqrt{5}}{3}\) is a frequent slip; always decide the sign from the quadrant before writing the value.
101IBMA3S-3-19r31-083.6 Trigonometric identitiesPaper 1Medium5 marks
No calculator
The vertical position of a probe's antenna is described by \(h=2\sin\left(2x-\frac{\pi}{3}\right)\) for \(0\le x\le\pi\). Find the values of \(x\) for which \(h=\sqrt{3}\). Give your answers as exact multiples of \(\pi\).
M1 states the range \(-\frac{\pi}{3}\le 2x-\frac{\pi}{3}\le\frac{5\pi}{3}\)
✓ 1
A1 \(2x-\frac{\pi}{3}=\frac{\pi}{3}\)
✓ 1
A1 \(2x-\frac{\pi}{3}=\frac{2\pi}{3}\)
✓ 1
A1 \(x=\frac{\pi}{3}\) and \(x=\frac{\pi}{2}\)
✓ 1
Exam tip: Convert the domain for x into a domain for the whole bracket before listing solutions, so that none are missed or added in error.
102IBMA3S-3-18r40-083.8 Trigonometric equationsPaper 1Medium6 marks
No calculator
Solve the equation \(2\sin^2 x = 1 + \cos x\) for \(0 \le x \le 2\pi\).
(a)
Give all the solutions in exact form.
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use \(\sin^2 x = 1 - \cos^2 x\)
✓ 1
A1 \(2\cos^2 x + \cos x - 1 = 0\)
✓ 1
M1 attempt to factorise: \((2\cos x - 1)(\cos x + 1) = 0\)
✓ 1
A1 \(\cos x = \frac{1}{2}\) or \(\cos x = -1\)
✓ 1
A1 \(x = \frac{\pi}{3}\) and \(x = \frac{5\pi}{3}\)
✓ 1
A1 \(x = \pi\)
✓ 1
Common error: Do not divide both sides by \(\cos x\) or cancel a common factor that could be zero: you lose solutions. Rearrange to zero and factorise.
103IBMA3S-3-18r40-143.6 Trigonometric identitiesPaper 1Medium10 marks
No calculator
This question is about the double-angle identities for sine and cosine.
(a)
Show that \(\frac{\sin 2\theta}{1 + \cos 2\theta} \equiv \tan\theta\).
M1 right-angled triangle with opposite \(1\), adjacent \(2\), hypotenuse \(\sqrt{5}\), so \(\cos\theta = \frac{2}{\sqrt{5}}\)
✓ 1
M1 \(\cos 2\theta = 2\cos^2\theta - 1\)
✓ 1
A1 \(2 \times \frac{4}{5} - 1 = \frac{3}{5}\)
✓ 1
Exam tip: Show-that questions need every step written: quote the identity you use, then simplify until you reach the printed result.
104IBMA3S-3-17r48-073.8 Trigonometric equationsPaper 1Medium6 marks
No calculator
Solve the equation \(2\sin^2 x + \sin x - 1 = 0\) for \(0 \le x \le 2\pi\), giving exact answers.
(a)
Solve the equation.
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 recognise a quadratic in \(\sin x\)
✓ 1
M1 attempt to factorise or solve: \((2\sin x - 1)(\sin x + 1) = 0\)
✓ 1
A1 \(\sin x = \frac{1}{2}\) or \(\sin x = -1\)
✓ 1
A1 \(x = \frac{\pi}{6}\)
✓ 1
A1 \(x = \frac{5\pi}{6}\)
✓ 1
A1 \(x = \frac{3\pi}{2}\)
✓ 1
Common error: \(\sin x = -1\) gives exactly one solution in \([0, 2\pi]\), and the solutions of \(\sin x = \frac{1}{2}\) come from \(x\) and \(\pi - x\).
105AASL3-q00463.5 The unit circle and the ambiguous casePaper 1Hard8 marks
In triangle \(ABC\), \(AB=10\) cm, \(BC=6\) cm and \(B\hat{A}C=30^{\circ}\).
(a)
Show that \(\sin A\hat{C}B=\dfrac{5}{6}\).
(2)
(b)
Explain why there are two possible triangles \(ABC\) that satisfy this information.
(2)
(c)
By using the cosine rule, find the two possible exact values of \(AC\).
\(A\hat{C}B\) may be acute (\(C_{1}\)) or obtuse, \(C_{2}=180^{\circ}-C_{1}\), since both have sine \(\dfrac{5}{6}\)
R1
since \(\dfrac{5}{6}\gt\dfrac{1}{2}=\sin30^{\circ}\), \(C_{1}\gt30^{\circ}\), so \(C_{2}+30^{\circ}\lt180^{\circ}\) and the obtuse case gives a valid triangle
\(6^{2}=10^{2}+x^{2}-2(10)x\cos30^{\circ}\), where \(x=AC\)
M1
\(x^{2}-10\sqrt{3}\,x+64=0\)
A1
\(x=\dfrac{10\sqrt{3}\pm\sqrt{300-256}}{2}\)
M1
Correct substitution into the quadratic formula, or completing the square.
\(AC=5\sqrt{3}+\sqrt{11}\) or \(AC=5\sqrt{3}-\sqrt{11}\)
A1
Both values required.
Answers: (c) \(AC=5\sqrt{3}\pm\sqrt{11}\) cm
106AASL3-q00543.5 The unit circle and the ambiguous casePaper 1Hard13 marks
In triangle \(ABC\), \(AB=12\) cm, \(B\hat{A}C=30^{\circ}\), \(BC=x\) cm and \(AC=y\) cm.
(a)
Show that \(y^{2}-12\sqrt3\,y+144-x^{2}=0\).
(2)
(b)
By considering this equation as a quadratic in \(y\), find the set of values of \(x\) for which two different triangles \(ABC\) can be drawn.
(4)
(c)
For a particular value of \(x\), two different triangles can be drawn and the length of \([AC]\) in one triangle is twice the length of \([AC]\) in the other. Find this value of \(x\).
(4)
(d)
For the value of \(x\) found in part (c), find the two possible sizes of \(A\hat{C}B\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(x^{2}=12^{2}+y^{2}-2(12)(y)\cos30^{\circ}\)
M1
\(\cos30^{\circ}=\dfrac{\sqrt3}{2}\), so \(x^{2}=144+y^{2}-12\sqrt3\,y\)
A1
\(y^{2}-12\sqrt3\,y+144-x^{2}=0\)
AG
(b)
two distinct roots: \(\left(12\sqrt3\right)^{2}-4\left(144-x^{2}\right)\gt0\)
M1
\(4x^{2}\gt144\), so \(x\gt6\)
A1
since \(x\gt0\)
both roots must be positive: their sum is \(12\sqrt3\gt0\), so their product \(144-x^{2}\gt0\)
R1
\(6\lt x\lt12\)
A1
(c)
roots \(k\) and \(2k\): \(3k=12\sqrt3\)
M1
\(k=4\sqrt3\)
A1
so \(AC=4\sqrt3\) or \(8\sqrt3\)
product of roots \(2k^{2}=144-x^{2}\), i.e. \(96=144-x^{2}\)
M1
Or substitute \(y=4\sqrt3\) into the equation in (a).
\(x=4\sqrt3\)
A1
Accept \(\sqrt{48}\). Award A0 if \(-4\sqrt3\) is also given.
(d)
valid method for one triangle, e.g. \(AC=BC=4\sqrt3\), so the triangle is isosceles with \(A\hat{B}C=30^{\circ}\); or \(AB^{2}+BC^{2}=144+48=192=AC^{2}\)
(M1)
Or sine rule: \(\sin A\hat{C}B=\dfrac{12\sin30^{\circ}}{4\sqrt3}=\dfrac{\sqrt3}{2}\).
\(\cos x\lt0\) and \(\sin x\gt0\) for \(\dfrac{\pi}{2}\lt x\lt\pi\), so \(\cos x-\sin x\lt0\)
R1
Alternative R1: \(\sin x+\cos x\gt0\) with \(\cos x\lt0\) gives \(\dfrac{\pi}{2}\lt x\lt\dfrac{3\pi}{4}\), so \(\pi\lt2x\lt\dfrac{3\pi}{2}\) and \(\cos2x\lt0\).
\(\cos x-\sin x=-\dfrac{\sqrt{7}}{2}\)
A1
In the alternative method, award for \(\cos2x=\pm\dfrac{\sqrt{7}}{4}\).
M1 integrates using \(f(x)=1-\cos 4x\): \(\int_0^{\pi/4}(1-\cos 4x)\,dx\)
✓ 1
A1 \(\left[x-\frac14\sin 4x\right]_0^{\pi/4}\)
✓ 1
A1 \(\frac{\pi}{4}\)
✓ 1
Exam tip: Rewriting a product of powers as a single cosine of a multiple angle makes range, period, equations and integration all straightforward.
114IBMA2S-048-02-123.5 The unit circle and the ambiguous casePaper 1Hard14 marks
No calculator
An engineer designs a triangular bracket ABC with AB = 8 cm and angle BAC = π/6. The length BC = k cm can be adjusted.
(a)
Find the set of values of k for which two distinct triangles ABC exist.
(3)
(b)
Let k = 4√2. Find the two possible values of angle ACB.
(4)
(c)
With k = 4√2, let AC = b. Show that b2 − 8√3b + 32 = 0.
(2)
(d)
Hence find the two possible exact values of b.
(2)
(e)
Find the ratio of the larger possible area of the bracket to the smaller, giving your answer in the form p + √q, where p, q ∈ ℤ.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 perpendicular distance from B to line AC is 8 sin(π/6)
✓ 1
A1 distance = 4
✓ 1
A1 4 < k < 8
✓ 1
Part (b)
M1 sine rule: sin C / 8 = sin(π/6) / (4√2)
✓ 1
A1 sin C = 4/(4√2) = √2/2
✓ 1
A1 C = π/4
✓ 1
A1 C = 3π/4
✓ 1
Part (c)
M1 cosine rule: 32 = 64 + b2 − 2(8)b cos(π/6)
✓ 1
A1 32 = 64 + b2 − 8√3 b, leading to the given equation
✓ 1
Part (d)
M1 b = [8√3 ± √(192 − 128)]/2
✓ 1
A1 b = 4√3 + 4 or 4√3 − 4
✓ 1
Part (e)
M1 area = ½ × 8 × b × sin(π/6) = 2b, so ratio = (4√3 + 4)/(4√3 − 4)
✓ 1
M1 rationalises (√3 + 1)/(√3 − 1) by multiplying by (√3 + 1)/(√3 + 1)
✓ 1
A1 (4 + 2√3)/2 = 2 + √3
✓ 1
Exam tip: The two values of b come from the same quadratic as the two angles from the sine rule: both describe the ambiguous case, so each value of C pairs with one value of b.
115IBMA2S-048-03-113.8 Trigonometric equationsPaper 1Hard14 marks
No calculator
Let \(f(x) = \cos 2x + k\sin x\), where \(k \in \mathbb{R}\) and \(0 \le x < 2\pi\).
(a)
Show that the equation \(f(x) = 0\) can be written as \(2\sin^2 x - k\sin x - 1 = 0\).
(2)
(b)
For \(k = 1\), solve the equation \(f(x) = 0\).
(4)
(c)
Let \(s = \sin x\). Show that, for all real k, the equation \(2s^2 - ks - 1 = 0\) has two distinct real roots, one positive and one negative.
(3)
(d)
Find the set of values of k for which the equation \(f(x) = 0\) has exactly four solutions.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(\cos 2x = 1 - 2\sin^2 x\)
✓ 1
A1 \(1 - 2\sin^2 x + k\sin x = 0 \Rightarrow 2\sin^2 x - k\sin x - 1 = 0\)
✓ 1
Part (b)
M1 factorises: \((2\sin x + 1)(\sin x - 1) = 0\)
✓ 1
A1 \(\sin x = 1\) or \(\sin x = -\frac{1}{2}\)
✓ 1
A1 \(x = \frac{\pi}{2}\)
✓ 1
A1 \(x = \frac{7\pi}{6}, \frac{11\pi}{6}\)
✓ 1
Part (c)
M1 discriminant \(= k^2 + 8\)
✓ 1
R1 \(k^2 + 8 > 0\) for all k, so there are two distinct real roots
✓ 1
R1 product of roots \(= -\frac{1}{2} < 0\), so the roots have opposite signs
✓ 1
Part (d)
R1 each value of s with \(-1 < s < 1\) gives two values of x, \(s = \pm 1\) gives one value, and \(|s| > 1\) gives none
✓ 1
R1 so both roots must lie strictly between −1 and 1
✓ 1
M1 lets \(g(s) = 2s^2 - ks - 1\); since \(g(0) = -1 < 0\), this requires \(g > 0\) and \(g(-1) > 0\)
✓ 2
A1 \(1 - k > 0\) and \(1 + k > 0\)
✓ 1
A1 \(-1 < k < 1\)
✓ 1
Exam tip: Count the solutions for x by working with the roots in s. Your result from part (b) gives a check: at \(k = 1\) there are exactly three solutions, so \(k = 1\) must be excluded.
116IBMA2S-048-00-123.1 Three-dimensional geometryPaper 1Hard13 marks
No calculator
A closed, hollow glass cone stands on its circular base on a table. Its interior is a right circular cone with base radius 6 cm and vertical height 8 cm. A solid ball (sphere) inside the cone rests on the base and touches the curved surface all the way round. Consider the vertical cross-section through the axis of the cone.
Vertical cross-section through the axis, not to scale
(a)
Find the slant height of the interior of the cone.
(2)
(b)
Let the ball have radius ρ cm. Show that ρ = 3.
(3)
(c)
Find the fraction of the interior volume of the cone occupied by the ball.
(4)
(d)
A second, smaller ball is placed above the first. It touches the first ball and the curved surface of the cone all the way round. Find the exact radius of the smaller ball.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 √(62 + 82)
✓ 1
A1 10 cm
✓ 1
Part (b)
M1 sine of the half-angle at the vertex = 6/10 = 3/5
✓ 1
M1 the centre is 8 − ρ from the vertex and ρ from the slant side, so ρ = (8 − ρ) × 3/5
✓ 1
A1 5ρ = 24 − 3ρ, so 8ρ = 24 and ρ = 3
✓ 1
(Accept any equivalent valid method, e.g. inradius of the isosceles cross-section = area ÷ semi-perimeter = 48/16 = 3.)
—
Part (c)
M1 cone volume (1/3)π(62)(8)
✓ 1
A1 96π
✓ 1
M1 ball volume (4/3)π(33) = 36π
✓ 1
A1 fraction = 36π/96π = 3/8
✓ 1
Part (d)
M1 top of the first ball is 6 cm above the base, so 2 cm below the vertex
✓ 1
M1 EITHER the region above the horizontal tangent plane at that point is a cone similar to the original with height 2 cm, and the smaller ball is the ball inscribed in it (it touches that plane at the top of the first ball) OR the centre of the smaller ball is 2 − r from the vertex
✓ 1
M1 EITHER scale factor 2/8 = 1/4 OR r = (3/5)(2 − r)
✓ 1
A1 r = 3/4 cm
✓ 1
Exam tip: A sphere touching a cone all round appears as a circle tangent to both slant sides in the cross-section. The radius drawn to the point of contact is perpendicular to the side, which gives a right-angled triangle.
117IBMA2S-048-01-113.8 Trigonometric equationsPaper 1Hard13 marks
No calculator
Let \(f(x)=2\cos(kx)\) for \(0\le x\le\pi\), where \(k>0\).
(a)
Let \(k=3\). (i) Write down the period of \(f\). (ii) Solve \(f(x)=1\).
(4)
(b)
Find the set of values of \(k\) for which the equation \(f(x)=1\) has exactly four solutions.
(4)
(c)
Let \(k=4\) and let \(g(x)=f\left(x-\frac{\pi}{8}\right)\). Show that \(g(x)=2\sin(4x)\).
(2)
(d)
Hence solve \(g(x)=f(x)\) for \(0\le x\le\pi\), with \(k=4\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
(i) A1 \(\frac{2\pi}{3}\)
✓ 1
(ii) M1 \(\cos(3x)=\frac{1}{2}\) with \(0\le 3x\le 3\pi\)
M1 solutions of \(\cos\theta=\frac{1}{2}\) for \(\theta\ge 0\): \(\frac{\pi}{3},\,\frac{5\pi}{3},\,\frac{7\pi}{3},\,\frac{11\pi}{3},\,\frac{13\pi}{3},\ldots\) with \(0\le\theta\le k\pi\)
✓ 1
R1 four solutions requires \(k\pi\ge\frac{11\pi}{3}\)
✓ 1
R1 and \(k\pi<\frac{13\pi}{3}\)
✓ 1
A1 \(\frac{11}{3}\le k<\frac{13}{3}\)
✓ 1
Part (c)
M1 \(g(x)=2\cos\left(4x-\frac{\pi}{2}\right)\)
✓ 1
A1 using \(\cos\left(\theta-\frac{\pi}{2}\right)=\sin\theta\), so \(g(x)=2\sin(4x)\)
Common error: In (b), the endpoint condition decides strict or non-strict inequalities. k = 11/3 gives exactly four solutions, but k = 13/3 gives a fifth solution at x = π.
118IBMA2S-048-06-123.2 Right-angled and non-right-angled trigonometryPaper 1Hard10 marks
No calculator
A folding phone stand has two struts hinged at H. One strut has length \(x\) cm and the other has length \((x+1)\) cm. When the stand is open, the angle between the struts is 120° and their free ends are joined by a straight bar of length \(\sqrt{37}\) cm.
Diagram not to scale
(a)
Show that \(x^2 + x - 12 = 0\).
(3)
(b)
Find the value of \(x\), giving a reason for rejecting any other solution.
(2)
(c)
Find the exact area of the triangle formed by the two struts and the bar.
(2)
(d)
Find the exact perpendicular distance from the hinge H to the bar, giving your answer in the form \(\frac{a\sqrt{b}}{c}\) where \(a, b, c \in \mathbb{Z}^+\).
A1 use of \(\cos 120° = -\frac{1}{2}\): \(37 = x^2 + (x+1)^2 + x(x+1)\)
✓ 1
A1 \(3x^2 + 3x - 36 = 0\) leading to \(x^2 + x - 12 = 0\) AG
✓ 1
Part (b)
A1 \((x+4)(x-3) = 0\) so \(x = 3\)
✓ 1
R1 \(x = -4\) rejected because a length must be positive
✓ 1
Part (c)
M1 \(\frac{1}{2}(3)(4)\sin 120°\)
✓ 1
A1 area = \(3\sqrt{3}\) cm2
✓ 1
Part (d)
M1 \(\frac{1}{2}\sqrt{37}\,d = 3\sqrt{3}\)
✓ 1
A1 \(d = \frac{6\sqrt{3}}{\sqrt{37}}\)
✓ 1
A1 \(d = \frac{6\sqrt{111}}{37}\) cm
✓ 1
Exam tip: The area of a triangle can be written in two ways: \(\frac{1}{2}ab\sin C\) and \(\frac{1}{2}\times\text{base}\times\text{height}\). Equating them is the quickest route to a perpendicular distance without finding any further angles.
119IBMA2S-048-04-123.4 Radian measure, arcs and sectorsPaper 1Hard12 marks
No calculator
A two-link robotic arm lies in a vertical plane. The shoulder joint O is fixed. The upper link OA has length 40 cm, and the forearm AB has length 24 cm, where B is the gripper. The elbow angle OAB is θ, where 0° < θ ≤ 180°.
Diagram not to scale
(a)
Show that \(OB^2 = 2176 - 1920\cos\theta\).
(2)
(b)
The gripper must reach a point 56 cm from O. Find the value of θ.
(3)
(c)
With the gripper 56 cm from O, find the exact value of sin(∠AOB).
(2)
(d)
With the gripper 56 cm from O, find the exact area of triangle OAB.
(2)
(e)
A different task requires triangle OAB to have the largest possible area. Find this maximum area, and find the exact distance OB in this position.
A1 1600 + 576 = 2176 and 2(40)(24) = 1920, giving the result
✓ 1
Part (b)
M1 \(3136 = 2176 - 1920\cos\theta\)
✓ 1
A1 \(\cos\theta = -\frac{1}{2}\)
✓ 1
A1 θ = 120°
✓ 1
Part (c)
M1 sine rule: \(\sin(\angle AOB) = \frac{24\sin 120°}{56}\)
✓ 1
A1 \(\sin(\angle AOB) = \frac{3\sqrt{3}}{14}\)
✓ 1
Part (d)
M1 Area \(= \frac{1}{2}(40)(24)\sin 120°\)
✓ 1
A1 Area \(= 240\sqrt{3}\) cm²
✓ 1
Part (e)
R1 area \(= 480\sin\theta\), which is greatest when sin θ = 1, so θ = 90°
✓ 1
A1 maximum area = 480 cm²
✓ 1
A1 \(OB = \sqrt{2176} = 8\sqrt{34}\) cm
✓ 1
Common error: Assuming the area is greatest when the arm is fully stretched (θ = 180°). The triangle then collapses to a straight line with zero area. The factor sin θ is largest at 90°.
120IBMA3S-3-17r6-143.6 Trigonometric identitiesPaper 1Hard7 marks
No calculator
It is given that \(\sin x+\cos x=\dfrac12\).
(a)
Show that \(\sin2x=-\dfrac34\).
(3)
(b)
Find the exact value of \(\cos4x\).
(2)
(c)
Hence find the exact value of \(\sin^3x+\cos^3x\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 square both sides: \(\sin^2x+2\sin x\cos x+\cos^2x=\frac14\)
✓ 1
M1 use \(\sin^2x+\cos^2x=1\)
✓ 1
A1 \(1+\sin2x=\frac14\) so \(\sin2x=-\frac34\) (AG)
✓ 1
Part (b)
M1 use \(\cos4x=1-2\sin^22x\)
✓ 1
A1 \(1-2\left(\frac9{16}\right)=-\frac18\)
✓ 1
Part (c)
M1 \((\sin x+\cos x)(1-\sin x\cos x)\) with \(\sin x\cos x=-\frac38\)
Exam tip: Squaring a sum is a quick route to a double-angle identity: the cross term gives 2 sin x cos x.
121IBMA3S-3-17r8-133.8 Trigonometric equationsPaper 1Hard7 marks
No calculator
The angle \(x\) (in radians) of a ship's crane arm satisfies \(\cos 2x=3\sin x-1\) for \(0\le x\le 2\pi\).
(a)
Show that the equation can be written as \(2\sin^2x+3\sin x-2=0\).
(3)
(b)
Hence solve the equation, giving exact values of \(x\).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use \(\cos2x=1-2\sin^2x\)
✓ 1
M1 substitute: \(1-2\sin^2x=3\sin x-1\)
✓ 1
A1 rearrange to \(2\sin^2x+3\sin x-2=0\) (AG)
✓ 1
Part (b)
M1 factorise \((2\sin x-1)(\sin x+2)=0\)
✓ 1
A1 \(\sin x=\frac12\) or \(\sin x=-2\)
✓ 1
R1 reject \(\sin x=-2\) because \(-1\le\sin x\le1\)
✓ 1
A1 \(x=\frac{\pi}{6}\) and \(x=\frac{5\pi}{6}\)
✓ 1
Common error: Do not forget to reject quadratic roots outside the range of sine or cosine; the equation may factorise but still give only one valid value of the function.
122IBMA3S-3-16r18-123.6 Trigonometric identitiesPaper 1Hard10 marks
No calculator
In this question, \(0<\theta<\pi\) and \(\theta\ne\frac{\pi}{2}\).
(a)
Show that \(\frac{1-\cos2\theta}{\sin2\theta}=\tan\theta\).
M1 writes the expression as \(2c^2-3c-1\) with \(c=\cos x\)
✓ 1
A1 vertex at \(c=\frac{3}{4}\), which lies in \([-1,1]\)
✓ 1
A1 minimum value \(=-\frac{17}{8}\)
✓ 1
Common error: After forming a quadratic in \(\cos x\), check each root lies between −1 and 1 before using it; for part (c) also check that the vertex is inside that interval.
125IBMA3S-3-17r48-143.4 Radian measure, arcs and sectorsPaper 1Hard7 marks
No calculator
A flatbread is cut into a sector of a circle with radius \(r\) cm and angle \(\theta\) radians, where \(0 < \theta \le 2\pi\). The perimeter of the sector is 24 cm and its area is 20 cm².
Diagram not to scale
(a)
Write down an equation for the perimeter and an equation for the area, in terms of \(r\) and \(\theta\).
(2)
(b)
Show that \(r^2 - 12r + 20 = 0\).
(2)
(c)
Hence find the value of \(r\) and the value of \(\theta\), explaining why one solution of the quadratic must be rejected.
R1 reject \(r = 2\) because it gives \(\theta = 10 > 2\pi\)
✓ 1
A1 \(r = 10\), \(\theta = 0.4\)
✓ 1
Exam tip: Always check solutions of a quadratic against the conditions in the question, here \(\theta \le 2\pi\).
126AASL3-q00103.1 Three-dimensional geometryPaper 2Easy9 marks
A fuel storage tank is in the shape of a right circular cylinder with a hemisphere attached at each end. The cylinder has radius \(3.8\) m and length \(13.8\) m, and each hemisphere has radius \(3.8\) m.
Diagram not to scale
(a)
Find the volume of the tank.
(3)
(b)
Find the external surface area of the tank.
(3)
(c)
The outside of the tank is to be painted with one coat of paint. One litre of paint covers \(12\text{ m}^{2}\) and the paint is sold in \(5\)-litre tins. Find the least number of tins needed.
(3)
Show mark scheme
Working
Mark
Notes
(a)
cylinder: \(\pi(3.8)^{2}(13.8)\;(=626.03)\)
(M1)
two hemispheres form a sphere: \(\dfrac{4}{3}\pi(3.8)^{3}\;(=229.85)\)
\(87.0636\ldots\) Accept \(87.4\) from using \(13.2\).
Answers: (a) \(13.2\) cm · (b) \(87.1\text{ cm}^{2}\)
128AASL3-q00203.2 Right-angled and non-right-angled trigonometryPaper 2Easy7 marks
Building regulations state that a wheelchair ramp must rise no more than \(1\) m vertically for every \(12\) m horizontally. A straight ramp is to be built from the ground to a doorway \(0.75\) m above the ground.
Diagram not to scale
(a)
Write down the least horizontal distance the ramp must cover.
(1)
(b)
Find the greatest angle that the ramp may make with the horizontal.
(2)
(c)
Find the length of the sloping surface of the ramp in part (a).
(2)
(d)
A builder proposes a straight ramp whose sloping surface is \(8.5\) m long, from the ground to the doorway. Determine whether this ramp meets the regulations.
\(5.06209\ldots^{\circ}\). Or \(0.0884\) radians. Or horizontal distance \(\sqrt{8.5^{2}-0.75^{2}}=8.47\lt9\).
\(5.06^{\circ}\gt4.76^{\circ}\), so the ramp is too steep and does not meet the regulations
R1
Do not award R1 without a correct comparison. Award (M1)R0 for a correct angle with an incorrect conclusion.
Answers: (a) \(9\) m · (b) \(4.76^{\circ}\) · (c) \(9.03\) m · (d) No (\(5.06^{\circ}\gt4.76^{\circ}\))
129AASL3-q00253.3 Applications of trigonometryPaper 2Easy5 marks
A vertical flagpole \([AB]\) of height \(7\) m stands on horizontal ground, with its base at \(B\). From a point \(C\) on the ground the angle of elevation of the top of the flagpole, \(A\), is \(32^{\circ}\). From a point \(D\) on \([CB]\) the angle of elevation of \(A\) is \(55^{\circ}\).
(a)
Find the distance \(BC\).
(2)
(b)
Find the distance \(CD\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\tan32^{\circ}=\dfrac{7}{BC}\)
(M1)
\(BC=11.2\) m
A1
\(11.2023\ldots\)
(b)
\(\tan55^{\circ}=\dfrac{7}{BD}\)
(M1)
\(BD=4.90\) m
(A1)
\(4.90145\ldots\)
\(CD=11.202\ldots-4.901\ldots=6.30\) m
A1
\(6.30088\ldots\) Accept \(6.3\). FT from (a).
Answers: (a) \(11.2\) m · (b) \(6.30\) m
130AASL3-q00293.3 Applications of trigonometryPaper 2Easy5 marks
The banks of a river are straight and parallel. Two points \(P\) and \(Q\) are \(30\) m apart on one bank. A tree \(T\) stands on the opposite bank. It is measured that \(T\hat{P}Q=75^{\circ}\) and \(T\hat{Q}P=50^{\circ}\).
Perpendicular distance from \(T\) to the line \((PQ)\). FT from (b).
\(27.1\) m
A1
\(27.0990\ldots\)
Answers: (a) \(55^{\circ}\) · (b) \(35.4\) m · (c) \(27.1\) m
131AASL3-q00413.4 Radian measure, arcs and sectorsPaper 2Easy7 marks
Diagram not to scale
A windscreen wiper rotates about the point \(O\) through an angle of \(2.2\) radians. The rubber blade lies along the arm, from \(12\) cm to \(54\) cm from \(O\). The shaded region shows the area of the windscreen wiped by the blade.
(a)
Find the area wiped by the blade.
(2)
(b)
Find the perimeter of the wiped region.
(3)
(c)
The wiper takes \(0.75\) seconds to rotate through the \(2.2\) radians. Find the average speed of the tip of the blade, in cm s−1.
the other angle with this sine is \(180^{\circ}-27.9^{\circ}=152.1^{\circ}\)
A1
\(152.1^{\circ}+40^{\circ}\gt 180^{\circ}\), so this angle is impossible
R1
Accept: the side opposite the given angle (11 cm) is longer than \(PQ\), so \(P\hat{R}Q\) must be acute.
(c)
\(P\hat{Q}R=180^{\circ}-40^{\circ}-27.87^{\circ}=112.1^{\circ}\) and \(\dfrac{PR}{\sin112.1^{\circ}}=\dfrac{11}{\sin40^{\circ}}\)
(M1)
\(15.9\) cm
A1
\(15.8523\ldots\)
Answers: (a) \(27.9^{\circ}\) · (c) 15.9 cm
133IBMA2S-048-05-033.4 Radian measure, arcs and sectorsPaper 2Easy5 marks
Calculator
The illuminance, \(E\) lux, at the centre of a desk from an adjustable lamp is modelled by \(E=400\sin\theta\cos\theta\), where \(\theta\) is the angle in radians between the lamp arm and the desk, \(0<\theta<\frac{\pi}{2}\).
(a)
Show that \(E=200\sin 2\theta\).
(2)
(b)
Find both values of \(\theta\) for which \(E=150\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(\sin 2\theta=2\sin\theta\cos\theta\), so \(\sin\theta\cos\theta=\frac{1}{2}\sin 2\theta\)
A1 \(2\theta=\pi-0.848\ldots=2.29\ldots\), so \(\theta=1.15\)
✓ 1
Common error: Solving for \(2\theta\) and then forgetting to halve, or finding only one solution, loses marks. The range \(0<2\theta<\pi\) contains two solutions.
134IBMA2S-048-07-023.4 Radian measure, arcs and sectorsPaper 2Easy5 marks
Calculator
In a hammer-throw competition the landing area is a sector of a circle centred on the middle of the throwing circle. The sector has a central angle of 34.92°. Distance lines are painted as circular arcs centred on the same point.
Plan view, not to scale
(a)
Convert 34.92° to radians.
(1)
(b)
Find the length of the painted arc that is 70 m from the centre.
(2)
(c)
Find the area of the part of the landing sector that lies between the 60 m arc and the 70 m arc.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 0.609 rad (0.609469…)
✓ 1
Part (b)
M1 s = 70 × 0.609469…
✓ 1
A1 42.7 m
✓ 1
Part (c)
M1 ½ × 0.609469… × (70² − 60²)
✓ 1
A1 396 m²
✓ 1
Common error: The formulas s = rθ and A = ½r²θ only work with θ in radians. Using 34.92 directly gives answers that are about 57 times too large.
135IBMA2S-048-07-043.4 Radian measure, arcs and sectorsPaper 2Easy4 marks
Calculator
Each blade of a wind turbine is 52 m long, measured from the centre of the hub. The blades rotate at a constant rate of 14 revolutions per minute.
(a)
Find the angle, in radians, that a blade turns through in one second.
(2)
(b)
Hence find the distance travelled by the tip of a blade in one second.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 14 × 2π ÷ 60
✓ 1
A1 7π/15 ≈ 1.47 rad
✓ 1
Part (b)
M1 s = 52 × 7π/15
✓ 1
A1 76.2 m
✓ 1
Exam tip: One revolution is 2π radians. Convert revolutions per minute to radians per second before you use s = rθ.
136IBMA2S-048-03-023.5 The unit circle and the ambiguous casePaper 2Easy5 marks
Calculator
A space station orbits the Earth. Its latitude, φ degrees, is modelled by \(\varphi(t) = 51.6\sin\left(\frac{2\pi t}{92.7}\right)\), where t is the time in minutes after the station crosses the equator heading north. Positive values of φ are north of the equator.
(a)
Find the first time after t = 0 at which the station is at latitude 30° north.
(3)
(b)
Find the next time at which the station is again at latitude 30° north.
Exam tip: Put the calculator in radian mode whenever t appears with π inside the sine. Before you finish, check that your two times are symmetric about a quarter-period, \(\frac{92.7}{4} = 23.2\) min.
137IBMA2S-048-03-063.6 Trigonometric identitiesPaper 2Easy4 marks
Calculator
The mains voltage, V volts, supplied to a laptop charger is modelled by \(V = 325\sin(100\pi t)\), where t is the time in seconds. Find the first time after t = 0 at which V = −200.
(a)
Find the first positive value of t for which V = −200.
Common error: The calculator returns \(\sin^{-1}(-0.615) = -0.663\), which gives a negative time. The first positive solution lies in the third quadrant, \(\pi + 0.663\).
138IBMA2S-048-00-023.1 Three-dimensional geometryPaper 2Easy5 marks
Calculator
A medical capsule is modelled as a cylinder with a hemisphere of the same radius at each end. The capsule has total length 21 mm and diameter 7 mm.
Diagram not to scale
(a)
Find the volume of the capsule.
(3)
(b)
The capsule is given a thin coating over its entire outer surface. Find the area to be coated.
Exam tip: Two hemispheres of equal radius combine into one full sphere, which halves the working. Remember the cylinder length is the total length minus one full diameter.
139IBMA2S-048-00-043.1 Three-dimensional geometryPaper 2Easy5 marks
Calculator
A spherical fuel tank on a satellite has an outer surface area of 15.0 m2. Ignore the thickness of the tank wall.
(a)
Find the radius of the tank.
(2)
(b)
Find the volume of the tank.
(2)
(c)
At launch the tank is 85% full. Find the volume of fuel in litres, where 1 m3 = 1000 litres.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 4πr2 = 15.0
✓ 1
A1 r = 1.09 m (1.0925…)
✓ 1
Part (b)
M1 (4/3)π(1.0925…)3 using the unrounded radius
✓ 1
A1 5.46 m3
✓ 1
Part (c)
A1 0.85 × 5.4627… = 4.643 m3, so 4640 litres
✓ 1
Exam tip: Keep the full calculator value of the radius when finding the volume. Cubing the rounded value 1.09 gives 5.42 m³, which is wrong to 3 significant figures.
140IBMA2S-048-01-073.4 Radian measure, arcs and sectorsPaper 2Easy7 marks
Calculator
A cyclist's pedal moves in a circle about the crank axle. The crank length is 17 cm and the axle is 28 cm above the ground. The cyclist pedals at a constant 80 revolutions per minute. At time \(t=0\) seconds the pedal is at its lowest point. The height of the pedal above the ground, \(h\) cm, is modelled by \(h(t)=a\cos(bt)+d\), where \(b>0\).
(a)
Write down the maximum height of the pedal above the ground.
(1)
(b)
Find the values of \(a\), \(b\) and \(d\).
(4)
(c)
Find the height of the pedal when \(t=2\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 45 cm
✓ 1
Part (b)
M1 period \(=\frac{60}{80}=0.75\) s
✓ 1
A1 \(b=\frac{2\pi}{0.75}=\frac{8\pi}{3}\) (8.38)
✓ 1
A1 \(a=-17\)
✓ 1
A1 \(d=28\)
✓ 1
Part (c)
M1 \(h(2)=28-17\cos\left(\frac{16\pi}{3}\right)\)
✓ 1
A1 36.5 cm
✓ 1
Common error: Starting at the lowest point means the cosine model must be reflected, so a is negative. Using a = +17 places the pedal at the top at t = 0.
141IBMA2S-048-06-013.3 Applications of trigonometryPaper 2Easy5 marks
Calculator
A laptop is opened so that the screen makes an angle of 115° with the keyboard base. The base is 30 cm deep (from the hinge to the front edge) and the screen is 22 cm tall (from the hinge to the top edge). Treat the base and the screen as two sides of a triangle that meet at the hinge.
Diagram not to scale
(a)
Find the straight-line distance from the front edge of the base to the top edge of the screen.
(3)
(b)
Find the area of the triangle formed by the base, the screen and this line.
M1 use of \(\frac{1}{2}ab\sin C\): \(\frac{1}{2}(30)(22)\sin 115°\)
✓ 1
A1 area = 299 cm2
✓ 1
Common error: Because 115° is obtuse, \(\cos 115°\) is negative, so the term \(-2ab\cos C\) is added. The third side must be longer than \(\sqrt{30^2+22^2}\approx 37.2\) cm; if your answer is shorter than this, check the sign of the cosine.
142IBMA2S-048-06-033.3 Applications of trigonometryPaper 2Easy5 marks
Calculator
Two tracking stations, A and B, are on level ground 12 km apart. A high-altitude research balloon, H, is in the same vertical plane as A and B and lies between them horizontally. The angle of elevation of H is 72° from A and 81° from B.
Diagram not to scale
(a)
Find the distance AH.
(3)
(b)
Find the height of the balloon above the ground.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 angle AHB = 180° − 72° − 81° = 27°
✓ 1
M1 sine rule with correct pairing: \(\frac{AH}{\sin 81°} = \frac{12}{\sin 27°}\)
✓ 1
A1 AH = 26.1 km
✓ 1
Part (b)
M1 a correct method for the height, e.g. \(h = AH\sin 72°\), or \(h = BH\sin 81°\), or \(\frac{h}{\tan 72°} + \frac{h}{\tan 81°} = 12\)
✓ 1
A1 \(h = 24.8\) km (follow through from their AH)
✓ 1
Common error: The side AH is opposite the angle at B, not the angle at A. Pair each side with the angle directly across from it before substituting into the sine rule.
143IBMA2S-048-06-043.6 Trigonometric identitiesPaper 2Easy4 marks
Calculator
A football goal is 7.32 m wide. A player stands so that she is 18 m from the left post and 21 m from the right post.
Diagram not to scale
(a)
Find the angle, at the player, between the lines joining her to the two posts. This is the angle within which a shot must travel to pass between the posts.
Exam tip: When all three sides are known, the side opposite the required angle goes in the subtracted term. Keep the full calculator value of \(\cos\theta\) before applying inverse cosine to avoid rounding errors.
144IBMA2S-048-04-013.2 Right-angled and non-right-angled trigonometryPaper 2Easy7 marks
Calculator
A wheelchair ramp is built to reach a doorway 0.75 m above level ground. The ramp surface is a straight sloping plane, and building regulations state that its angle of inclination to the horizontal must not exceed 4.8°. The ramp is built at exactly 4.8°.
Diagram not to scale
(a)
Find the horizontal distance covered by the ramp.
(2)
(b)
Find the length of the sloping surface of the ramp.
(2)
(c)
At a second building, the doorway is also 0.75 m high but the ramp has a horizontal distance of 10.0 m. Find the angle of inclination of this ramp and determine whether it meets the regulation.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 uses \(\tan 4.8° = \frac{0.75}{d}\)
✓ 1
A1 d = 8.93 m
✓ 1
Part (b)
M1 uses \(\sin 4.8° = \frac{0.75}{L}\), or Pythagoras with the answer to (a)
✓ 1
A1 L = 8.96 m
✓ 1
Part (c)
M1 \(\tan\theta = \frac{0.75}{10.0}\)
✓ 1
A1 θ = 4.29°
✓ 1
R1 4.29° < 4.8°, so the ramp meets the regulation
✓ 1
Exam tip: Identify the sides relative to the angle before choosing sin, cos or tan: the 0.75 m rise is opposite the angle, the horizontal run is adjacent and the ramp surface is the hypotenuse.
145IBMA2S-048-04-033.2 Right-angled and non-right-angled trigonometryPaper 2Easy5 marks
Calculator
A camera drone hovers at a fixed height of 120 m above level ground, filming a cyclist. At first, the angle of depression from the drone to the cyclist is 32°. The cyclist then rides in a straight line directly towards the point on the ground vertically below the drone, and the angle of depression becomes 51°.
Diagram not to scale
(a)
Find the initial horizontal distance between the cyclist and the point vertically below the drone.
(2)
(b)
Find the distance the cyclist rides between the two observations.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\tan 32° = \frac{120}{d_1}\) or equivalent
✓ 1
A1 \(d_1\) = 192 m (192.04…)
✓ 1
Part (b)
M1 \(d_2 = \frac{120}{\tan 51°}\) or equivalent
✓ 1
A1 \(d_2\) = 97.2 m (97.17…)
✓ 1
M1 subtracts: distance = 192.04 − 97.17 = 94.9 m (accept 94.8 m from 3 s.f. intermediate values)
✓ 1
Exam tip: The angle of depression from the drone equals the angle of elevation from the cyclist (alternate angles). You can therefore place it at ground level in the right-angled triangle. Keep full calculator values until the final subtraction to avoid rounding errors.
146IBMA2S-048-04-043.3 Applications of trigonometryPaper 2Easy6 marks
Calculator
In an orienteering race, a runner leaves the start S and runs 850 m on a bearing of 060° to checkpoint P. She then runs 1200 m on a bearing of 150° to checkpoint Q.
Diagram not to scale
(a)
Show that angle SPQ = 90°.
(1)
(b)
Find the distance SQ.
(2)
(c)
Find the bearing of Q from S.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 the bearing of S from P is 240°, and 240° − 150° = 90°, so angle SPQ = 90°
✓ 1
Part (b)
M1 \(SQ^2 = 850^2 + 1200^2\)
✓ 1
A1 SQ = 1470 m
✓ 1
Part (c)
M1 \(\tan(\angle PSQ) = \frac{1200}{850}\)
✓ 1
A1 angle PSQ = 54.7°
✓ 1
A1 bearing = 60° + 54.7° = 115°
✓ 1
Common error: Giving angle PSQ (54.7°) as the final answer. A bearing is measured clockwise from north, so the 60° of the first leg must be added.
147IBMA3S-3-17r6-013.5 The unit circle and the ambiguous casePaper 2Easy4 marks
Calculator
An aircraft climbs in a straight line at a constant angle of 12° to the horizontal. It flies 8.5 km along its flight path. Give each answer to 3 significant figures.
Diagram not to scale
(a)
Find the gain in altitude.
(2)
(b)
Find the horizontal distance covered.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use \(8.5\sin 12^\circ\)
✓ 1
A1 1.77 km
✓ 1
Part (b)
M1 attempt to use \(8.5\cos 12^\circ\)
✓ 1
A1 8.31 km
✓ 1
Common error: Set the calculator to degrees for this question; a radian-mode answer will look plausible but be wrong.
148IBMA3S-3-17r6-023.2 Right-angled and non-right-angled trigonometryPaper 2Easy6 marks
Calculator
Two microphones, A and B, stand on a level stage 6.0 m apart. A singer stands at S. Angle SAB = 52° and angle SBA = 71°. Give lengths to 3 significant figures.
M1 attempt to use \(AS\sin 52^\circ\) (ECF from (b))
✓ 1
A1 5.33 m
✓ 1
Exam tip: Find the third angle first; the sine rule needs a side opposite a known angle.
149IBMA3S-3-17r6-033.4 Radian measure, arcs and sectorsPaper 2Easy4 marks
Calculator
A rotary sprinkler in a field waters a sector of a circle with radius 18 m. The sector angle at the sprinkler is 2.4 radians.
Diagram not to scale
(a)
Find the length of the curved boundary of the watered region.
(2)
(b)
Find the area watered. Give your answer to the nearest m2.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use \(s=r\theta\)
✓ 1
A1 43.2 m
✓ 1
Part (b)
M1 attempt to use \(\frac12 r^2\theta\)
✓ 1
A1 388.8 = 389 m2
✓ 1
Common error: The formulas \(s=r\theta\) and \(A=\frac12r^2\theta\) only work when \(\theta\) is in radians.
150IBMA3S-3-17r8-013.3 Applications of trigonometryPaper 2Easy5 marks
Calculator
A surveyor stands on level ground 850 m horizontally from the foot of a vertical cliff on the rim of a volcanic crater. Her eye is at ground level. The angle of elevation of the top of the cliff is 32°.
Diagram not to scale
(a)
Find the height of the top of the cliff above the ground. Give your answer to the nearest metre.
(2)
(b)
She walks towards the cliff until the angle of elevation of the top is 45°. Use the unrounded height from (a). Find how far she walks, to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use \(\tan 32^\circ=\frac{h}{850}\)
✓ 1
A1 \(h=850\tan 32^\circ=531\) m
✓ 1
Part (b)
M1 at 45° the horizontal distance equals the height (\(\tan45^\circ=1\)) so distance is \(531.14\ldots\) m
✓ 1
M1 subtract from 850: \(850-531.14\ldots\)
✓ 1
A1 \(318.86\ldots\) so 319 m
✓ 1
Common error: Keep the unrounded value from part (a) on your calculator; rounding to 531 first would change the third significant figure here.
151IBMA3S-3-17r8-023.4 Radian measure, arcs and sectorsPaper 2Easy4 marks
Calculator
The floor of a theatre stage is a sector of a circle with radius 7.2 m. The angle at the centre of the sector is 1.9 radians.
Diagram not to scale
(a)
Find the length of the curved edge of the stage, to 3 significant figures.
(2)
(b)
Find the area of the stage floor, to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(s=r\theta\) with \(r=7.2\), \(\theta=1.9\)
✓ 1
A1 \(13.68=13.7\) m
✓ 1
Part (b)
M1 use of \(A=\frac12 r^2\theta\)
✓ 1
A1 \(\frac12(7.2)^2(1.9)=49.248=49.2\) m\(^2\)
✓ 1
Common error: The formulas \(s=r\theta\) and \(A=\frac12r^2\theta\) need \(\theta\) in radians; do not switch the calculator to degrees for this question.
152IBMA3S-3-17r8-033.5 The unit circle and the ambiguous casePaper 2Easy4 marks
Calculator
A cargo ship leaves port and sails 18 km due east, then 24 km due north.
Diagram not to scale
(a)
Find the direct distance from the port to the ship's final position.
(2)
(b)
Find the bearing of the ship from the port, to 1 decimal place.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of Pythagoras \(\sqrt{18^2+24^2}\)
✓ 1
A1 30 km
✓ 1
Part (b)
M1 attempt to use \(\tan\alpha=\frac{18}{24}\) for the angle from north
✓ 1
A1 \(\alpha=36.9^\circ\), so bearing \(036.9^\circ\)
✓ 1
Exam tip: Bearings are measured clockwise from north; sketch the two legs and mark the angle from the north line before choosing a trig ratio.
153IBMA3S-3-16r18-013.4 Radian measure, arcs and sectorsPaper 2Easy4 marks
Calculator
A blade of a wind turbine has length 40 m and rotates about the centre of the hub. In one gust the blade turns through 1.8 radians.
Diagram not to scale
(a)
Find the distance travelled by the tip of the blade.
(2)
(b)
Find the area swept out by the blade during this turn.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of arc length \(s=r\theta\)
✓ 1
A1 \(s=40\times1.8=72\) m
✓ 1
Part (b)
M1 use of sector area \(A=\frac{1}{2}r^2\theta\)
✓ 1
A1 \(A=\frac{1}{2}\times40^2\times1.8=1440\) m2
✓ 1
Common error: The formulas \(s=r\theta\) and \(A=\frac{1}{2}r^2\theta\) only work when the angle is in radians.
154IBMA3S-3-16r18-033.3 Applications of trigonometryPaper 2Easy4 marks
Calculator
A ladder of length 3.6 m rests against a library bookshelf. The ladder makes an angle of 68° with the horizontal floor. Give your answers to 3 significant figures.
Diagram not to scale
(a)
Find the height above the floor at which the ladder touches the bookshelf.
(2)
(b)
Find the horizontal distance from the foot of the ladder to the bookshelf.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(h=3.6\sin68^\circ\)
✓ 1
A1 \(h=3.34\) m
✓ 1
Part (b)
M1 \(d=3.6\cos68^\circ\)
✓ 1
A1 \(d=1.35\) m
✓ 1
Exam tip: Sketch the right-angled triangle and label the hypotenuse first so you pick sine or cosine correctly.
155IBMA3S-3-17r24-013.3 Applications of trigonometryPaper 2Easy4 marks
Calculator
A volcanologist stands at a point P on level ground, 3100 m horizontally from the base of a volcano. She measures the angle of elevation of the summit from P as 14.5°. Her eyes are at ground level.
Diagram not to scale
(a)
Find the height of the summit above P, to the nearest metre.
(2)
(b)
Find the straight-line distance from P to the summit, to the nearest metre.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of tan 14.5° = h/3100
✓ 1
A1 h = 801.7 = 802 m
✓ 1
Part (b)
M1 use of cos 14.5° = 3100/d (or Pythagoras with the unrounded height)
✓ 1
A1 d = 3202.0 = 3202 m
✓ 1
Common error: Check that the calculator is in degree mode before using an angle given in degrees; a wrong mode gives an answer that looks plausible but is far off.
156IBMA3S-3-17r24-043.3 Applications of trigonometryPaper 2Easy4 marks
Calculator
Two navigation buoys A and B are 640 m apart. A ship S is positioned so that angle SAB = 48° and angle SBA = 67°.
Diagram not to scale
(a)
Find the size of angle ASB.
(1)
(b)
Find the distance from the ship to buoy A, to the nearest metre.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 180° − 48° − 67° = 65°
✓ 1
Part (b)
M1 sine rule with the side AB opposite the 65° angle
✓ 1
A1 SA/sin 67° = 640/sin 65°
✓ 1
A1 SA = 650.0 = 650 m
✓ 1
Exam tip: In the sine rule, pair each side with the angle directly opposite it; label the triangle first so you do not mix them.
157IBMA3S-3-17r25-013.2 Right-angled and non-right-angled trigonometryPaper 2Easy5 marks
Calculator
Three beehives, A, B and C, stand in a flat field. AB = 48 m, AC = 65 m and angle BAC = 72°.
Diagram not to scale
(a)
Calculate the distance BC, giving your answer to 3 significant figures.
(3)
(b)
Calculate the area of triangle ABC, giving your answer to 3 significant figures.
M1 attempt to use area = ½ab sin C, e.g. ½(48)(65)sin72°
✓ 1
A1 area = 1480 m2
✓ 1
Exam tip: When you are given two sides and the angle between them, the cosine rule finds the third side and the formula ½ab sin C gives the area directly.
158IBMA3S-3-17r25-033.8 Trigonometric equationsPaper 2Easy4 marks
Calculator
Solve the equation 4 sin x − 3 = 0 for 0 ≤ x ≤ 2π, giving your answers in radians correct to 3 significant figures.
(a)
Solve the equation.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 rearrange to sin x = \(\frac{3}{4}\)
✓ 1
A1 x = 0.848
✓ 1
M1 use symmetry: second solution is π − (first solution)
✓ 1
A1 x = 2.29
✓ 1
Exam tip: On a sine equation over 0 to 2π, a positive value of sin x always gives two solutions: the calculator value and π minus that value.
159IBMA3S-3-17r25-043.2 Right-angled and non-right-angled trigonometryPaper 2Easy5 marks
Calculator
A hospital entrance has a straight ramp of length 9.5 m that rises through a vertical height of 0.80 m.
Diagram not to scale
(a)
Calculate the angle the ramp makes with the horizontal, giving your answer to 3 significant figures.
(2)
(b)
The hospital's design guide states that the angle of a ramp must not exceed 4.5°. State whether this ramp meets the guide, giving a reason.
(1)
(c)
The rise must stay 0.80 m. Calculate the least length of ramp that meets the guide, giving your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use sin θ = \(\frac{0.80}{9.5}\)
✓ 1
A1 θ = 4.83°
✓ 1
Part (b)
A1 does not meet the guide, since 4.83° > 4.5° (ECF from (a))
✓ 1
Part (c)
M1 use sin 4.5° = \(\frac{0.80}{L}\)
✓ 1
A1 L = 10.2 m
✓ 1
Exam tip: In a right-angled triangle, label opposite, adjacent and hypotenuse first; here the rise is opposite the angle and the ramp is the hypotenuse, so use sine.
160IBMA3S-3-19r27-013.3 Applications of trigonometryPaper 2Easy4 marks
Calculator
A cable car runs in a straight line from a base station up to a ridge station. The ridge station is directly above a point on the ground that is 1450 m horizontally from the base station. The angle of elevation of the ridge station from the base station is 28°.
Diagram not to scale
(a)
Find the height of the ridge station above the base station, to the nearest metre.
(2)
(b)
Find the length of the cable, to the nearest metre.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 tan 28° = h/1450, or h = 1450 tan 28°
✓ 1
A1 h = 771 m
✓ 1
Part (b)
M1 cos 28° = 1450/L, or Pythagoras with 1450 and the height
✓ 1
A1 L = 1642 m
✓ 1
Common error: Check your calculator is in degree mode before using tan 28°; a radian-mode answer is wildly different.
161IBMA3S-3-19r27-033.4 Radian measure, arcs and sectorsPaper 2Easy5 marks
Calculator
A round cheese wheel of radius 14 cm is cut so that one slice is a sector with an angle of 75° at the centre of the wheel.
Diagram not to scale
(a)
Write 75° in radians, as an exact multiple of \(\pi\).
(1)
(b)
Find the length of the curved edge of the slice, in cm to 3 significant figures.
(2)
(c)
Find the area of the top face of the slice, in cm2 to 3 significant figures.
Exam tip: The formulae \(s=r\theta\) and \(A=\frac12r^2\theta\) need \(\theta\) in radians; convert first or use the fraction \(\frac{75}{360}\) of the circle.
162IBMA3S-3-19r27-043.2 Right-angled and non-right-angled trigonometryPaper 2Easy4 marks
Calculator
The cross-section of a bar of nougat is a triangle. Two sides of the triangle have lengths 5.5 cm and 6.0 cm, and the angle between them is 52°.
Diagram not to scale
(a)
Find the area of the cross-section, in cm2 to 3 significant figures.
(2)
(b)
Find the length of the third side, in cm to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac12\times5.5\times6.0\times\sin52°\)
✓ 1
A1 13.0 cm2
✓ 1
Part (b)
M1 \(c^2=5.5^2+6.0^2-2(5.5)(6.0)\cos52°\)
✓ 1
A1 5.06 cm
✓ 1
Exam tip: Use the area formula with the angle that lies between the two given sides; the cosine rule then gives the side opposite that angle.
163IBMA3S-3-19r30-013.3 Applications of trigonometryPaper 2Easy4 marks
Calculator
A climber fixes a taut rope of length 85 m from a point on level ground at the foot of a snow slope to an anchor on a rock ledge. The rope makes an angle of 62° with the horizontal ground. Give answers to 3 significant figures.
Diagram not to scale
(a)
Find the height of the ledge above the ground.
(2)
(b)
Find the horizontal distance from the foot of the rope to the point on the ground directly below the anchor.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 h = 85 sin 62°
✓ 1
A1 h = 75.1 m
✓ 1
Part (b)
M1 d = 85 cos 62°
✓ 1
A1 d = 39.9 m
✓ 1
Common error: Check that your calculator is in degree mode before using sin, cos or tan with an angle given in degrees.
164IBMA3S-3-19r30-033.4 Radian measure, arcs and sectorsPaper 2Easy5 marks
Calculator
A mobile-phone mast at O broadcasts to a sector-shaped region of radius 9 km. The sector has angle 1.4 radians at O.
Diagram not to scale
(a)
Find the length of the curved boundary of the region.
(2)
(b)
Find the area of the region.
(2)
(c)
Find the total perimeter of the region.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 arc length = 9 × 1.4
✓ 1
A1 12.6 km
✓ 1
Part (b)
M1 area = ½ × 9² × 1.4
✓ 1
A1 56.7 km²
✓ 1
Part (c)
A1 2 × 9 + 12.6 = 30.6 km
✓ 1
Exam tip: The formulas s = rθ and A = ½r²θ only work when the angle is in radians.
165IBMA3S-3-19r30-043.7 Circular functions and their graphsPaper 2Easy6 marks
Calculator
The strength S of the signal received from a rotating antenna is modelled by S = 55 + 8 sin(20t)°, where t is the time in minutes and 0 ≤ t ≤ 18.
(a)
Write down the maximum and the minimum value of S.
(2)
(b)
Write down the period of the model.
(1)
(c)
Find the values of t for which S = 61, giving answers to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 maximum 63
✓ 1
A1 minimum 47
✓ 1
Part (b)
A1 360 ÷ 20 = 18 minutes
✓ 1
Part (c)
M1 sin(20t) = 0.75
✓ 1
A1 t = 2.43
✓ 1
A1 t = 6.57 (from 20t = 180° − 48.59…°)
✓ 1
Common error: A sine equation has a second solution in each period; subtract the calculator angle from 180° to find it.
166IBMA3S-3-19r31-013.3 Applications of trigonometryPaper 2Easy6 marks
Calculator
Two ground antennas, A and B, are 1200 m apart on level ground. They both track a landing probe P that is in the same vertical plane as A and B. Angle PAB = 62° and angle PBA = 74°.
Diagram not to scale
(a)
Find angle APB.
(1)
(b)
Find the distance AP, giving your answer to the nearest metre.
(3)
(c)
Hence find the height of P above the line AB, to the nearest metre. Use the unrounded value of AP.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 angle APB = 180° − 62° − 74° = 44°
✓ 1
Part (b)
M1 attempt to use the sine rule with AP opposite angle B
Common error: Checking the calculator is in degree mode is worth doing before any trigonometry; a radian-mode answer here would be wildly wrong.
167IBMA3S-3-19r31-033.3 Applications of trigonometryPaper 2Easy6 marks
Calculator
A bridge deck is horizontal. A straight stay cable of length 85 m runs from a point on the deck up to a point T on a vertical tower. The cable makes an angle of 38° with the deck.
Diagram not to scale
(a)
Find the height of T above the deck, correct to 3 significant figures.
(2)
(b)
Find the horizontal distance from the foot of the tower to the point where the cable meets the deck, correct to 3 significant figures.
(2)
(c)
A second cable runs from the same point on the deck to a point 20 m higher up the tower than T. Find the angle this second cable makes with the deck. Use unrounded values from parts (a) and (b), and give your answer to the nearest 0.1°.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 height = 85 sin 38°
✓ 1
A1 52.3 m
✓ 1
Part (b)
M1 distance = 85 cos 38°
✓ 1
A1 67.0 m
✓ 1
Part (c)
M1 tan θ = (85 sin 38° + 20)/(85 cos 38°)
✓ 1
A1 θ = 47.2°
✓ 1
Common error: When a later part builds on earlier answers, carry the unrounded values forward and round only at the end.
168IBMA3S-3-18r40-013.3 Applications of trigonometryPaper 2Easy5 marks
Calculator
A cable-stayed bridge has a vertical tower with its top at point \(T\) and its foot at point \(F\) on a horizontal deck. Points \(A\) and \(B\) are on the deck in a straight line with \(F\), with \(A\) between \(F\) and \(B\), and \(AB = 80\) m. The angle of elevation of \(T\) from \(A\) is \(52^\circ\) and from \(B\) is \(38^\circ\).
Diagram not to scale
(a)
Find the height \(TF\) of the tower. Give your answer to 3 significant figures, using unrounded intermediate values.
M1 sine rule: \(\frac{AT}{\sin 38^\circ} = \frac{80}{\sin 14^\circ}\)
✓ 1
A1 \(AT = 203.6\ldots\) m
✓ 1
M1 \(TF = AT\sin 52^\circ\)
✓ 1
A1 \(TF = 160\) m
✓ 1
Exam tip: An angle of elevation from the farther point is smaller, so the angle at the top of the triangle is the difference of the two elevations.
169IBMA3S-3-18r40-023.4 Radian measure, arcs and sectorsPaper 2Easy4 marks
Calculator
A chute in a coffee roaster has a cross-section in the shape of a sector of a circle of radius \(r\) cm. The angle of the sector is \(1.4\) radians and its area is \(140\) cm\(^2\).
Diagram not to scale
(a)
Find the value of \(r\), correct to 3 significant figures.
(2)
(b)
Find the perimeter of the sector, using the unrounded value of \(r\). Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{1}{2}r^2(1.4) = 140\)
✓ 1
A1 \(r = 14.1\) cm
✓ 1
Part (b)
M1 perimeter \(= 2r + 1.4r\)
✓ 1
A1 \(48.1\) cm
✓ 1
Common error: The perimeter of a sector includes the two straight radii as well as the arc, so it is \(2r + r\theta\), not just \(r\theta\).
170IBMA3S-3-18r40-053.2 Right-angled and non-right-angled trigonometryPaper 2Easy5 marks
Calculator
A triangular steel brace in a bridge truss has sides of length \(7.2\) m, \(9.5\) m and \(12.0\) m.
Diagram not to scale
(a)
Find the size of the largest angle of the brace. Give your answer to 3 significant figures.
(3)
(b)
Find the area of the brace. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 cosine rule for the angle opposite the \(12.0\) m side
✓ 1
A1 \(\cos C = \frac{7.2^2 + 9.5^2 - 12.0^2}{2(7.2)(9.5)}\)
✓ 1
A1 \(C = 90.8^\circ\)
✓ 1
Part (b)
M1 area \(= \frac{1}{2}(7.2)(9.5)\sin C\)
✓ 1
A1 \(34.2\) m\(^2\)
✓ 1
Exam tip: The largest angle is opposite the longest side, so choose that side as the one on the left of the cosine rule.
171IBMA3S-3-17r48-013.3 Applications of trigonometryPaper 2Easy4 marks
Calculator
A cable car runs in a straight line from a base camp to a ridge. The angle of elevation of the ridge from the base camp is 28°, and the ridge is 640 m horizontally from the base camp. The ground below the cable is treated as a straight slope.
Diagram not to scale
(a)
Find the height of the ridge above the base camp, giving your answer to the nearest metre.
(2)
(b)
Find the length of the cable, giving your answer to the nearest metre. Use the data in the question, not your answer to part (a).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use \(\tan 28° = \frac{h}{640}\)
✓ 1
A1 \(h = 640\tan 28° = 340.3\), so height = 340 m
✓ 1
Part (b)
M1 attempt to use \(\cos 28° = \frac{640}{L}\)
✓ 1
A1 \(L = \frac{640}{\cos 28°} = 724.8\), so cable length = 725 m
✓ 1
Common error: Make sure the calculator is in degree mode before using the given angle; a radian-mode answer is wildly different.
172IBMA3S-3-17r48-093.7 Circular functions and their graphsPaper 2Easy6 marks
Calculator
A telecoms engineer models the voltage \(V\) volts of a signal at time \(t\) milliseconds by \(V = a\sin(bt) + c\), where \(a > 0\), \(b > 0\) and \(bt\) is in radians. The maximum voltage is 7 V, the minimum voltage is \(-1\) V, and the period is 4 ms.
(a)
Find the value of \(a\) and the value of \(c\).
(2)
(b)
Find the exact value of \(b\).
(2)
(c)
Find the voltage when \(t = 0.5\), to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(a = \frac{7-(-1)}{2} = 4\)
✓ 1
A1 \(c = \frac{7+(-1)}{2} = 3\)
✓ 1
Part (b)
M1 attempt to use period \(= \frac{2\pi}{b}\), so \(4 = \frac{2\pi}{b}\)
✓ 1
A1 \(b = \frac{\pi}{2}\)
✓ 1
Part (c)
M1 substitute into \(V = 4\sin\left(\frac{\pi t}{2}\right) + 3\) (ECF from (a) and (b))
✓ 1
A1 \(V = 3 + 2\sqrt{2} = 5.83\) V
✓ 1
Exam tip: For \(y = a\sin(bt) + c\), the amplitude is half the difference of the extremes and \(c\) is the midpoint.
173AASL3-q00063.1 Three-dimensional geometryPaper 2Medium8 marks
A solid metal sphere has volume \(500\text{ cm}^{3}\).
(a)
Find the radius of the sphere.
(2)
(b)
Find the surface area of the sphere.
(2)
(c)
The sphere is melted down and all of the metal is recast into a solid right circular cone. The radius of the base of the cone is equal to the radius, \(r\) cm, of the sphere. Show that the height of the cone is \(4r\) cm.
(2)
(d)
Find the curved surface area of the cone.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{4}{3}\pi r^{3}=500\)
(M1)
\(r=4.92\) cm
A1
\(4.92372\ldots\)
(b)
\(4\pi(4.92372\ldots)^{2}\)
(M1)
FT from their \(r\).
\(305\text{ cm}^{2}\)
A1
\(304.647\ldots\) Accept \(304\)–\(305\).
(c)
\(\dfrac{1}{3}\pi r^{2}h=\dfrac{4}{3}\pi r^{3}\)
M1
Equating the two volumes in terms of r. Award M0 for working only with 500 and a numerical r.
\(r^{2}h=4r^{3}\)
A1
\(h=4r\)
AG
(d)
slant height \(l=\sqrt{r^{2}+(4r)^{2}}=r\sqrt{17}\;(=20.3)\) and \(\pi rl\)
174AASL3-q00073.1 Three-dimensional geometryPaper 2Medium8 marks
Diagram not to scale
The diagram shows a right pyramid \(VABCD\) with a square base of side \(10\) cm. The point \(M\) is the centre of the base and \(V\) is vertically above \(M\), with \(VM=12\) cm.
(a)
Find the volume of the pyramid.
(2)
(b)
Find the angle between the edge \(VA\) and the base \(ABCD\).
175AASL3-q00093.1 Three-dimensional geometryPaper 2Medium8 marks
Diagram not to scale
A solid is made from a right circular cylinder of radius \(r\) cm and height \(3r\) cm, with a solid right circular cone of base radius \(r\) cm and height \(2r\) cm fixed to its top face, as shown in the diagram.
(a)
Show that the volume of the solid is \(\dfrac{11}{3}\pi r^{3}\text{ cm}^{3}\).
(2)
(b)
The volume of the solid is \(2000\text{ cm}^{3}\). Find the value of \(r\).
(2)
(c)
Show that the slant height of the cone is \(\sqrt{5}\,r\) cm.
(1)
(d)
Find the total surface area of the solid.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\pi r^{2}(3r)+\dfrac{1}{3}\pi r^{2}(2r)\)
M1
Sum of the volumes of the cylinder and the cone, both in terms of r.
\(3\pi r^{3}+\dfrac{2}{3}\pi r^{3}\)
A1
\(\dfrac{11}{3}\pi r^{3}\)
AG
(b)
\(\dfrac{11}{3}\pi r^{3}=2000\)
(M1)
\(r=5.58\)
A1
\(5.57874\ldots\)
(c)
\(l=\sqrt{r^{2}+(2r)^{2}}=\sqrt{5r^{2}}\)
A1
Must see the Pythagoras step.
\(l=\sqrt{5}\,r\)
AG
(d)
cylinder: base \(\pi r^{2}\) and curved surface \(2\pi r(3r)=6\pi r^{2}\)
(M1)
Award M0 if the top face of the cylinder (inside the solid) is included.
cone: curved surface \(\pi r(\sqrt{5}\,r)\); total \(=(7+\sqrt{5})\pi r^{2}\)
(M1)
\(903\text{ cm}^{2}\)
A1
\(903.045\ldots\) FT from (b). Accept \(902\)–\(904\).
176AASL3-q00113.1 Three-dimensional geometryPaper 2Medium9 marks
Diagram not to scale
The diagram shows a right pyramid \(VABCD\) with a horizontal rectangular base \(ABCD\), where \(AB=16\) cm and \(BC=10\) cm. The point \(M\) is the point of intersection of the diagonals of the base, and \(V\) is vertically above \(M\) with \(VM=9\) cm.
(a)
Find the length of the edge \([VA]\).
(2)
(b)
The point \(N\) is the midpoint of \([BC]\). Find the angle between the line \([VN]\) and the base \(ABCD\).
(2)
(c)
Find the size of the angle \(A\hat{V}C\).
(2)
(d)
Find the total surface area of the pyramid.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(AM=\dfrac{1}{2}\sqrt{16^{2}+10^{2}}=\sqrt{89}\;(=9.43)\) and \(VA=\sqrt{89+9^{2}}\)
(M1)
\(VA=\sqrt{170}=13.0\) cm
A1
\(13.0384\ldots\)
(b)
\(MN=8\) and \(\tan V\hat{N}M=\dfrac{9}{8}\)
(M1)
Or \(VN=\sqrt{145}\) and \(\sin V\hat{N}M=\dfrac{9}{\sqrt{145}}\). Must identify angle \(V\hat{N}M\); award M0 for \(MN=5\).
\(48.4^{\circ}\)
A1
\(48.3664\ldots^{\circ}\); \(0.844\) radians
(c)
\(A\hat{V}C=2\times A\hat{V}M\), where \(\tan A\hat{V}M=\dfrac{\sqrt{89}}{9}\)
(M1)
Or \(180^{\circ}-2\times V\hat{A}M\) with \(\tan V\hat{A}M=\dfrac{9}{\sqrt{89}}\), or the cosine rule in triangle \(AVC\) with \(AC=2\sqrt{89}\).
\(92.7^{\circ}\)
A1
\(92.6972\ldots^{\circ}\); \(1.62\) radians
(d)
heights of the triangular faces: \(\sqrt{9^{2}+5^{2}}=\sqrt{106}\) (faces on \(AB\), \(CD\)) and \(\sqrt{9^{2}+8^{2}}=\sqrt{145}\) (faces on \(BC\), \(DA\))
177AASL3-q00133.1 Three-dimensional geometryPaper 2Medium9 marks
A vertical pole \([OT]\) of height \(12\) m stands on horizontal ground. A coordinate system is set up with \(O\) at the origin, the \(x\)-axis pointing east, the \(y\)-axis pointing north and the \(z\)-axis vertically upwards; units are metres. The ground is the plane \(z=0\). Two straight ropes join the top of the pole, \(T\), to the points \(A(9,\,-4,\,0)\) and \(B(-3,\,8,\,0)\) on the ground.
(a)
Write down the coordinates of \(T\).
(1)
(b)
Find the length of the rope \([TA]\).
(2)
(c)
Find the angle between the rope \([TA]\) and the ground.
(2)
(d)
Find the coordinates of \(M\), the midpoint of \([AB]\).
(1)
(e)
Find the angle of elevation of \(T\) from \(M\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(T(0,\,0,\,12)\)
A1
(b)
\(\sqrt{9^{2}+(-4)^{2}+(-12)^{2}}\)
(M1)
\(\sqrt{241}=15.5\) m
A1
\(15.5241\ldots\)
(c)
\(OA=\sqrt{97}\) and \(\tan T\hat{A}O=\dfrac{12}{\sqrt{97}}\)
(M1)
Or \(\sin T\hat{A}O=\dfrac{12}{\sqrt{241}}\).
\(50.6^{\circ}\)
A1
\(50.6229\ldots^{\circ}\); \(0.884\) radians
(d)
\(M(3,\,2,\,0)\)
A1
(e)
\(OM=\sqrt{3^{2}+2^{2}}=\sqrt{13}\)
(A1)
FT from (d).
\(\tan T\hat{M}O=\dfrac{12}{\sqrt{13}}\)
(M1)
Or \(TM=\sqrt{157}=12.5\) and \(\sin T\hat{M}O=\dfrac{12}{\sqrt{157}}\).
Answers: (b) \(48.6^{\circ}\) or \(131^{\circ}\) · (c) \(16.4\) cm · (d) \(34.4\) cm
180AASL3-q00263.3 Applications of trigonometryPaper 2Medium6 marks
Diagram not to scale
A vertical cliff \([TF]\) is \(45\) m high, where \(F\) is at sea level. A boat sails in a straight line directly towards \(F\). At point \(A\) the angle of depression of the boat from \(T\) is \(12^{\circ}\). Two minutes later the boat is at point \(B\) and the angle of depression is \(20^{\circ}\).
(a)
Find the distance \(FA\).
(2)
(b)
Find the distance \(AB\).
(2)
(c)
Find the average speed of the boat between \(A\) and \(B\), giving your answer in km h−1.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\tan12^{\circ}=\dfrac{45}{FA}\)
(M1)
Angle of elevation of \(T\) from \(A\) is \(12^{\circ}\) (alternate angles).
\(FA=212\) m
A1
\(211.708\ldots\)
(b)
\(FB=\dfrac{45}{\tan20^{\circ}}\;(=123.636\ldots)\) and subtracting
(M1)
\(AB=88.1\) m
A1
\(88.0719\ldots\) FT from (a).
(c)
\(\dfrac{88.07\ldots}{120}=0.734\text{ m s}^{-1}\) or \(\dfrac{0.08807\ldots}{\tfrac{2}{60}}\)
(M1)
Correct conversion of both units.
\(2.64\text{ km h}^{-1}\)
A1
\(2.64215\ldots\) FT from (b).
Answers: (a) \(212\) m · (b) \(88.1\) m · (c) \(2.64\text{ km h}^{-1}\)
181AASL3-q00273.3 Applications of trigonometryPaper 2Medium8 marks
Two points \(A\) and \(B\) lie on horizontal ground, \(120\) m apart. A vertical tower \([FT]\) has its base \(F\) on the line segment \([AB]\). The angle of elevation of the top of the tower, \(T\), is \(28^{\circ}\) from \(A\) and \(41^{\circ}\) from \(B\).
Diagram not to scale
(a)
Write down the size of \(A\hat{T}B\).
(1)
(b)
Find \(AT\).
(2)
(c)
Find the height of the tower.
(2)
(d)
An observer walks from \(A\) directly towards the tower and stops at the midpoint of \([AB]\). Find the angle of elevation of \(T\) from this point.
Or \(\dfrac{h}{\tan28^{\circ}}+\dfrac{h}{\tan41^{\circ}}=120\).
\(39.6\) m
A1
\(39.5896\ldots\)
(d)
\(AF=\dfrac{39.58\ldots}{\tan28^{\circ}}=74.46\ldots\) or \(84.32\ldots\cos28^{\circ}\)
(M1)
horizontal distance from the midpoint to \(F\) \(=74.46\ldots-60=14.46\ldots\) and \(\tan\theta=\dfrac{39.58\ldots}{14.46\ldots}\)
(M1)
\(69.9^{\circ}\)
A1
\(69.9388\ldots^{\circ}\) FT from (c).
Answers: (a) \(111^{\circ}\) · (b) \(84.3\) m · (c) \(39.6\) m · (d) \(69.9^{\circ}\)
182AASL3-q00303.3 Applications of trigonometryPaper 2Medium11 marks
A vertical radio mast \([MT]\) of height \(40\) m stands on horizontal ground with its base at \(M\). Point \(A\) is \(30\) m from \(M\) on a bearing of \(030^{\circ}\), and point \(B\) is \(25\) m from \(M\) on a bearing of \(140^{\circ}\).
Plan view, not to scale
(a)
Find the angle of elevation of \(T\) from \(A\).
(2)
(b)
Find the distance \(AB\).
(3)
(c)
Find the bearing of \(B\) from \(A\).
(3)
(d)
A technician walks in a straight line from \(A\) to \(B\). Find the greatest angle of elevation of \(T\) from the technician during this walk.
183AASL3-q00403.4 Radian measure, arcs and sectorsPaper 2Medium7 marks
The diagram shows a sector \(OAB\) of a circle with centre \(O\) and radius \(r\) cm. The angle \(A\hat{O}B\) is \(\theta\) radians. The length of the arc \(AB\) is \(10\) cm and the area of the sector is \(35\text{ cm}^{2}\).
(a)
Find the value of \(r\) and the value of \(\theta\).
(4)
(b)
Find the area of the shaded segment.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(r\theta=10\) and \(\dfrac{1}{2}r^{2}\theta=35\)
(M1)
\(\dfrac{1}{2}r(r\theta)=35\Rightarrow5r=35\)
(M1)
Any valid method of eliminating one variable.
\(r=7\)
A1
\(\theta=\dfrac{10}{7}\;(=1.43)\)
A1
\(1.42857\ldots\)
(b)
area of triangle \(OAB=\dfrac{1}{2}(7)^{2}\sin\dfrac{10}{7}\;(=24.25)\)
(M1)
GDC must be in radian mode. Award M0 for \(\sin1.43^{\circ}\).
184AASL3-q00423.4 Radian measure, arcs and sectorsPaper 2Medium8 marks
A pendulum consists of a small bob on a light string of length \(80\) cm, attached to a fixed point \(O\). The bob swings in a vertical plane between two extreme positions \(A\) and \(B\). The angle \(A\hat{O}B\) is \(0.6\) radians, and the string is vertical when the bob is midway between \(A\) and \(B\).
Diagram not to scale
(a)
Find the length of the arc travelled by the bob in moving from \(A\) to \(B\).
(2)
(b)
Write down \(A\hat{O}B\) in degrees.
(1)
(c)
Find the straight-line distance \(AB\).
(2)
(d)
Find the height of \(A\) above the lowest point of the bob’s path.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(s=80\times0.6\)
(M1)
\(48\) cm
A1
(b)
\(34.4^{\circ}\)
A1
\(34.3774\ldots^{\circ}\)
(c)
\(AB=2\times80\sin0.3\)
(M1)
Or \(AB^{2}=80^{2}+80^{2}-2(80)(80)\cos0.6\). GDC in radian mode.
\(47.3\) cm
A1
\(47.2832\ldots\)
(d)
angle between \([OA]\) and the vertical is \(0.3\) radians
(M1)
vertical depth of \(A\) below \(O\) \(=80\cos0.3\;(=76.43)\); height \(=80-80\cos0.3\)
(M1)
\(3.57\) cm
A1
\(3.57308\ldots\)
Answers: (a) \(48\) cm · (b) \(34.4^{\circ}\) · (c) \(47.3\) cm · (d) \(3.57\) cm
185AASL3-q00523.5 The unit circle and the ambiguous casePaper 2Medium8 marks
A straight path starts at point \(A\). A post \(B\) is \(12\) m from \(A\), and \([AB]\) makes an angle of \(38^{\circ}\) with the path. A goat is tied to the post with a rope of length \(8.5\) m. When the rope is taut, the goat can reach the path at exactly two points, \(C_{1}\) and \(C_{2}\), as shown.
Diagram not to scale
(a)
Find the two possible sizes of the angle \(A\hat{C}B\), where \(C\) is a point on the path with \(BC=8.5\) m.
(3)
(b)
Find the distance \(C_{1}C_{2}\).
(3)
(c)
Find the least length of rope for which the goat can just reach the path.
\(119.637\ldots^{\circ}\). Accept 120°. FT \(180^{\circ}-\) their acute angle.
(b)
triangle \(BC_{1}C_{2}\) is isosceles with base angles \(60.36\ldots^{\circ}\), or \(AC_{1}=5.25\) and \(AC_{2}=13.7\) (cosine rule or sine rule)
(M1)
\(C_{1}C_{2}=2(8.5)\cos60.36\ldots^{\circ}\), or \(13.6595\ldots-5.25275\ldots\)
(M1)
\(8.41\) m
A1
\(8.40675\ldots\)
(c)
perpendicular distance from \(B\) to the path, \(12\sin38^{\circ}\)
(M1)
\(7.39\) m
A1
\(7.38793\ldots\)
Answers: (a) \(60.4^{\circ}\) and \(119.6^{\circ}\) · (b) 8.41 m · (c) 7.39 m
186AASL3-q00653.6 Trigonometric identitiesPaper 2Medium10 marks
The diagram shows a semicircle with centre \(O\) and radius \(10\) cm. The rectangle \(PQRS\) has \(P\) and \(Q\) on the diameter and \(R\) and \(S\) on the arc, and \(O\) is the midpoint of \([PQ]\). The angle \(Q\hat{O}R=\theta\), where \(0\lt\theta\lt\dfrac{\pi}{2}\).
Diagram not to scale
(a)
Show that the area of the rectangle is \(100\sin2\theta\text{ cm}^{2}\).
(3)
(b)
Write down the maximum area of the rectangle and the value of \(\theta\) at which it occurs.
(2)
(c)
Find the values of \(\theta\) for which the area of the rectangle is \(80\text{ cm}^{2}\).
(3)
(d)
Find the maximum perimeter of the rectangle.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(QR=10\sin\theta\) and \(OQ=10\cos\theta\)
A1
area \(=2(10\cos\theta)(10\sin\theta)=100(2\sin\theta\cos\theta)\)
M1
using \(\sin2\theta=2\sin\theta\cos\theta\)
A1
area \(=100\sin2\theta\)
AG
(b)
\(100\text{ cm}^{2}\)
A1
\(\theta=\dfrac{\pi}{4}\)
A1
Accept 0.785.
(c)
\(100\sin2\theta=80\), e.g. graph of \(y=100\sin2\theta\) with \(y=80\)
(M1)
\(\theta=0.464\)
A1
\(0.463647\ldots\)
\(\theta=1.11\)
A1
\(1.10714\ldots\)
(d)
perimeter \(=40\cos\theta+20\sin\theta\), maximum from GDC
(M1)
maximum at \(\theta=0.464\)
\(44.7\) cm
A1
\(44.7213\ldots=20\sqrt5\)
Answers: (b) 100 cm² at \(\theta=\frac{\pi}{4}\) · (c) 0.464, 1.11 · (d) 44.7 cm
187AASL3-q00743.7 Circular functions and their graphsPaper 2Medium8 marks
The depth of water, \(d\) metres, at the entrance to a harbour \(t\) hours after midnight is modelled by \(d(t)=2.6\cos\big(q(t-3)\big)+4.2\), for \(0\le t\le24\), where \(q\gt0\). The time between consecutive high tides is \(12.4\) hours.
(a)
Write down the depth of water at high tide.
(1)
(b)
Show that \(q=\dfrac{5\pi}{31}\).
(2)
(c)
Find the time of the first low tide after midnight, giving your answer to the nearest minute in 24-hour clock format.
(2)
(d)
A ship can enter the harbour only when the depth of water is at least \(3\) m. Find the total length of time between midnight and noon during which the ship can enter the harbour.
188AASL3-q00753.7 Circular functions and their graphsPaper 2Medium9 marks
The blood pressure, \(P\) mmHg, of a patient at rest is modelled by \(P(t)=98+22\sin(2.5\pi t)\), where \(t\) is the time in seconds, \(t\ge0\). Each period of \(P\) corresponds to one heartbeat.
(a)
Write down the maximum and the minimum blood pressure.
(2)
(b)
Find the number of heartbeats per minute.
(2)
(c)
Find the blood pressure when \(t=1.1\).
(2)
(d)
During exercise, the patient’s maximum and minimum blood pressures are each \(20\,\%\) higher than at rest, and the heart beats \(96\) times per minute. The blood pressure is then modelled by \(Q(t)=p+q\sin(kt)\), where \(p,q,k\gt0\). Find the value of \(p\), the value of \(q\) and the exact value of \(k\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
maximum \(120\) mmHg
A1
minimum \(76\) mmHg
A1
(b)
period \(=\dfrac{2\pi}{2.5\pi}=0.8\) s
(M1)
\(\dfrac{60}{0.8}=75\)
A1
(c)
\(98+22\sin(2.75\pi)\)
(M1)
\(114\) mmHg
A1
\(113.556\ldots\)
(d)
new maximum \(144\), new minimum \(91.2\); \(p=\dfrac{144+91.2}{2}=117.6\)
A1
\(q=\dfrac{144-91.2}{2}=26.4\)
A1
FT their (a).
period \(\dfrac{60}{96}=0.625\) s, so \(k=\dfrac{2\pi}{0.625}=3.2\pi\)
A1
Accept \(\dfrac{16\pi}{5}\). Award A0 for 10.1 only (exact value required).
189AASL3-q00763.7 Circular functions and their graphsPaper 2Medium8 marks
The mean monthly temperature, \(T\) °C, in a city is modelled using sinusoidal regression on a GDC. The calculator gives
\(T(x)=6.2\sin(0.524x-2.09)+14.8\),
where \(x\) is the month number (\(x=1\) for January, \(x=2\) for February, and so on), for \(1\le x\le12\).
(a)
Write \(T(x)\) in the form \(a\sin\big(b(x+c)\big)+d\), giving the value of \(c\) to three significant figures.
(2)
(b)
Find the period of \(T\) and comment on its value in this context.
(2)
(c)
Find the maximum mean monthly temperature predicted by the model, and the month in which it occurs.
(2)
(d)
Find the values of \(x\) for which \(T(x)=18\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(0.524\left(x-\dfrac{2.09}{0.524}\right)\)
(M1)
\(T(x)=6.2\sin\big(0.524(x-3.99)\big)+14.8\), so \(c=-3.99\)
A1
\(-3.98854\ldots\)
(b)
\(\dfrac{2\pi}{0.524}=12.0\)
A1
\(11.9908\ldots\)
approximately 12 months: the temperatures repeat every year
R1
(c)
\(21.0\) °C
A1
July
A1
maximum at \(x=6.99\) \((6.98625\ldots)\)
(d)
graph of \(T\) with \(y=18\), or \(\sin(0.524x-2.09)=\dfrac{3.2}{6.2}\)
(M1)
\(x=5.02\) and \(x=8.95\)
A1
\(5.02352\ldots,\ 8.94898\ldots\)
Answers: (a) \(c=-3.99\) · (b) 12.0 months · (c) 21.0 °C in July · (d) 5.02, 8.95
190AASL3-q00773.7 Circular functions and their graphsPaper 2Medium10 marks
The number of hours of daylight, \(D\), in a town on day \(t\) of the year is modelled by
\(D(t)=3.1\cos\left(\dfrac{2\pi}{365}(t-172)\right)+12.2\), for \(1\le t\le365,\ t\in\mathbb{Z}\),
where \(t=1\) is 1 January.
(a)
Write down the maximum number of hours of daylight and the value of \(t\) at which it occurs.
(2)
(b)
Find the number of hours of daylight on day 50.
(2)
(c)
Find the number of days in the year on which there are more than \(14\) hours of daylight.
(3)
(d)
In a second town, the longest day has \(16.8\) hours of daylight and the shortest day has \(7.6\) hours. These occur on the same days of the year as in the first town, and the second town is modelled by \(E(t)=a\cos\left(\dfrac{2\pi}{365}(t-172)\right)+d\). Find the number of days in the year on which the second town has more hours of daylight than the first town.
for \(x\gt8\), \(\dfrac{x}{2}\gt4\), so the two sides cannot be equal
R1
(c)
one further solution for \(2\pi\lt x\le8\), at \(x=6.63\)
(M1)
\(6.62810\ldots\)
\(5\)
A1
Answers: (a) 0.879, 2.39, 3.78, 5.85 · (c) 5
194IBMA2S-048-05-073.2 Right-angled and non-right-angled trigonometryPaper 2Medium11 marks
Calculator
A symmetric roof truss is an isosceles triangle whose two sloping beams are each 8 m long. The angle between each sloping beam and the vertical line of symmetry is \(\alpha\), so the apex angle is \(2\alpha\), where \(0<\alpha<\frac{\pi}{2}\).
Diagram not to scale
(a)
Show that the area of the truss, \(A\) m2, is given by \(A=32\sin 2\alpha\).
(3)
(b)
Write down the maximum possible area and the value of \(\alpha\) at which it occurs.
(2)
(c)
The truss must have area 28 m2 and \(\alpha>\frac{\pi}{4}\). Find \(\alpha\) and hence the height of the truss.
(4)
(d)
Find the length of the base of this truss.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 height \(=8\cos\alpha\) and base \(=16\sin\alpha\)
✓ 1
M1 \(A=\frac12\times16\sin\alpha\times8\cos\alpha=64\sin\alpha\cos\alpha\), or \(A=\frac12\times8\times8\sin2\alpha\)
✓ 1
A1 uses \(\sin 2\alpha=2\sin\alpha\cos\alpha\) (or the area formula directly) to reach \(A=32\sin 2\alpha\)
✓ 1
Part (b)
A1 maximum area 32 m2
✓ 1
A1 \(\alpha=\frac{\pi}{4}\)
✓ 1
Part (c)
M1 \(\sin 2\alpha=0.875\)
✓ 1
A1 \(2\alpha=\pi-1.0654\ldots=2.0762\ldots\), so \(\alpha=1.04\) (1.03808…)
✓ 1
M1 height \(=8\cos\alpha\), using the unrounded value of \(\alpha\)
✓ 1
A1 height \(=4.06\) m (4.0630…)
✓ 1
Part (d)
M1 base \(=16\sin\alpha\) with \(\alpha=1.03808\ldots\)
✓ 1
A1 13.8 m (13.783…)
✓ 1
Common error: The calculator gives \(2\alpha=1.07\), which leads to \(\alpha<\frac{\pi}{4}\). The condition requires the second solution \(\pi-1.07\). Carry full accuracy into later parts: using \(\alpha=1.04\) gives a height of 4.05 m instead of 4.06 m.
195IBMA2S-048-07-063.4 Radian measure, arcs and sectorsPaper 2Medium7 marks
Calculator
A security camera mounted at point C on a warehouse floor sweeps out a horizontal sector of radius r metres and central angle θ radians. The area of the floor covered by the camera is 180 m². The curved boundary of this region has length 24 m.
Plan view, not to scale
(a)
Find the value of r.
(3)
(b)
Find the value of θ.
(2)
(c)
Find the straight-line distance between the two ends of the curved boundary.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 writes ½r²θ = 180 and rθ = 24
✓ 1
M1 eliminates θ, e.g. ½r(rθ) = ½r × 24 = 180
✓ 1
A1 r = 15
✓ 1
Part (b)
M1 θ = 24/15
✓ 1
A1 θ = 1.6
✓ 1
Part (c)
M1 2 × 15 × sin(0.8), or √(15² + 15² − 2 × 15 × 15 cos 1.6)
✓ 1
A1 21.5 m
✓ 1
Exam tip: Dividing the area equation by the arc-length equation gives r/2 = 180/24 at once, which avoids solving simultaneous equations the long way.
196IBMA2S-048-07-083.4 Radian measure, arcs and sectorsPaper 2Medium12 marks
Calculator
A section of railway track follows a circular arc. The centre line of the track has radius 800 m, and the length of the curved section, measured along the centre line, is 1.2 km. The two rails are 1.435 m apart and lie symmetrically either side of the centre line, each on an arc with the same centre.
(a)
Find the angle, in radians, subtended at the centre of the circle by the curved section.
(2)
(b)
Find how much longer the outer rail is than the inner rail on this curve.
(3)
(c)
A train travels along the centre line of the curved section at a constant speed of 25 m s−1. Find the rate, in radians per second, at which the train turns about the centre of the circle.
(2)
(d)
Find the straight-line distance between the point where the centre line of the curved section starts and the point where it ends.
(2)
(e)
Find the area of the land enclosed between the centre line of the curve and the straight line joining its ends.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 θ = 1200/800
✓ 1
A1 θ = 1.5 rad
✓ 1
Part (b)
M1 radii 800.7175 m and 799.2825 m, OR radius difference = 1.435 m
✓ 1
M1 difference of arcs (800.7175 − 799.2825) × 1.5 or 1.435 × 1.5
✓ 1
A1 2.15 m (2.1525 m)
✓ 1
Part (c)
M1 time = 1200/25 = 48 s, OR ω = v/r = 25/800
✓ 1
A1 1.5/48 = 0.03125 ≈ 0.0313 rad s−1
✓ 1
Part (d)
M1 2 × 800 × sin(0.75), or cosine rule √(800² + 800² − 2 × 800² cos 1.5)
✓ 1
A1 1090 m (1091 m)
✓ 1
Part (e)
M1 sector area ½ × 800² × 1.5 = 480 000 m²
✓ 1
M1 subtracts triangle area ½ × 800² × sin 1.5 (with 1.5 in radians)
✓ 1
A1 160 802 ≈ 1.61 × 105 m²
✓ 1
Exam tip: For the difference in rail length, use (outer radius − inner radius) × θ = 1.435 × θ directly. This avoids the rounding errors that come from subtracting two large arc lengths.
197IBMA2S-048-07-093.4 Radian measure, arcs and sectorsPaper 2Medium7 marks
Calculator
When viewed from Earth, the Moon subtends an angle of 0.518° at the observer. The distance from the observer to the Moon is 384 400 km.
(a)
Convert 0.518° to radians.
(2)
(b)
By treating the Moon's diameter as an arc of a circle centred on the observer, estimate the diameter of the Moon.
(2)
(c)
A space station moves in a circular orbit of radius 6780 km about the centre of the Earth and completes one orbit every 92.7 minutes. Find the distance the space station travels in 10 minutes.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.518 × π/180
✓ 1
A1 0.00904 rad (0.0090408…)
✓ 1
Part (b)
M1 s = 384 400 × 0.0090408…
✓ 1
A1 3480 km (3475…)
✓ 1
Part (c)
M1 angle = 2π × 10/92.7 (= 0.678 rad)
✓ 1
M1 s = 6780 × that angle
✓ 1
A1 4600 km (4595…)
✓ 1
Exam tip: For very small angles, the arc and the chord are almost equal, so s = rθ gives an excellent estimate of a straight-line width seen from far away.
198IBMA2S-048-02-033.2 Right-angled and non-right-angled trigonometryPaper 2Medium6 marks
Calculator
A delivery drone's GPS logs three waypoints P, Q and R. The distance PQ is 8.4 km, the distance QR is 6.1 km and angle QPR = 38°. The distance PR was not recorded.
(a)
Find the two possible values of angle PRQ.
(3)
(b)
Given that angle PRQ is obtuse, find the distance PR.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sine rule: sin R / 8.4 = sin 38° / 6.1, so sin R = 0.8478
A1 PR = 3.38 km (accept 3.39 km from rounded angles)
✓ 1
Exam tip: Keep the unrounded value of angle R in your calculator when finding angle PQR; rounding to 122° first can shift the final answer in the third significant figure.
199IBMA2S-048-02-053.5 The unit circle and the ambiguous casePaper 2Medium6 marks
Calculator
The mainsail of a small yacht is a triangle with two edges of length 4.2 m and 5.5 m. The area of the sail is 9.0 m2. Let θ be the angle between the two given edges.
(a)
Find the two possible values of θ.
(3)
(b)
The third edge of the sail is known to be its longest edge. Find the length of the third edge.
R1 area = ½ × AB × BC × sin B is larger for B = 66.3° since sin 66.3° > sin 17.7°
✓ 1
M1 area = ½ × 9.2 × 7.5 × sin 66.27°
✓ 1
A1 31.6 cm2
✓ 1
Exam tip: Using the two known sides in the area formula requires the angle between them (angle B), so find B from the angle sum first.
201IBMA2S-048-03-053.7 Circular functions and their graphsPaper 2Medium6 marks
Calculator
A cyclist pedals at a constant cadence. The height, h metres, of one pedal above the road is modelled by \(h(t) = 0.30 - 0.17\cos(3\pi t)\), where t is the time in seconds.
(a)
Write down the maximum height of the pedal and the time taken for one revolution.
(2)
(b)
Find the length of time, during one revolution, for which the pedal is more than 0.40 m above the road.
M1 second boundary value \(3\pi t = 2\pi - 2.1997 = 4.0835\) (or uses symmetry)
✓ 1
A1 time \(= \frac{4.0835 - 2.1997}{3\pi} = 0.200\) s
✓ 1
Exam tip: Find both boundary times in one cycle and subtract them. A quick check is that the answer must be less than half of the period, \(\frac{1}{3}\) s, because 0.40 m is above the mean height.
202IBMA2S-048-03-093.7 Circular functions and their graphsPaper 2Medium7 marks
Calculator
A patient's body temperature, T °C, is modelled by \(T(t) = 36.8 + a\sin\left(\frac{\pi(t - 11)}{12}\right)\), where t is the number of hours after midnight, \(0 \le t < 24\), and a > 0. At 07:00 the patient's temperature is 36.5 °C.
(a)
Find the exact value of a.
(3)
(b)
Find the values of t at which the temperature is 37.0 °C.
Exam tip: Carry the exact value of a, or its full calculator value, into part (b). Using the rounded value 0.346 can change the third significant figure of your times.
203IBMA2S-048-03-103.7 Circular functions and their graphsPaper 2Medium14 marks
Calculator
The brightness of a variable star is measured on an arbitrary scale. Its brightness is modelled by \(B(t) = p + q\sin(rt)\), where t is the time in days and p, q, r > 0. The maximum brightness is 4.6, the minimum brightness is 3.4, and the period is 5.4 days.
(a)
Find the value of p and the value of q.
(2)
(b)
Find the exact value of r.
(2)
(c)
Find the exact value of the first time t > 0 at which B = 4.3.
(3)
(d)
A telescope can detect the star only when B > 3.7. Find the total time in one period for which the star can be detected.
(4)
(e)
Find the first two values of t > 0 for which B = 3.5.
Exam tip: As a check in part (e), the two times should be symmetric about the time of minimum brightness, which is three-quarters of a period: \(0.75 \times 5.4 = 4.05\) days.
204IBMA2S-048-00-063.1 Three-dimensional geometryPaper 2Medium7 marks
Calculator
A glass skylight is a right pyramid with a horizontal square base of side 2.4 m and its apex vertically above the centre of the base. Each triangular face has a perpendicular height (from the apex to the midpoint of a base edge) of 2.0 m.
Diagram not to scale
(a)
Find the vertical height of the skylight.
(2)
(b)
Find the total area of glass in the four triangular faces.
(2)
(c)
Find the angle between a sloping edge of the skylight and its base.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 h2 + 1.22 = 2.02
✓ 1
A1 h = 1.6 m
✓ 1
Part (b)
M1 4 × ½ × 2.4 × 2.0
✓ 1
A1 9.6 m2
✓ 1
Part (c)
M1 half-diagonal of base = 1.2√2 = 1.697… m
✓ 1
M1 tan θ = 1.6/1.697…
✓ 1
A1 θ = 43.3°
✓ 1
Common error: A sloping edge projects onto half the base diagonal, not half the side. Using 1.2 m here gives the angle of a face, 53.1°, instead of the edge.
205IBMA2S-048-00-073.1 Three-dimensional geometryPaper 2Medium6 marks
Calculator
A paper cup at a water cooler is a right circular cone with its vertex at the bottom. The rim has radius 4 cm and the cup has depth 9 cm. Water is poured in to a depth of h cm, and the water surface has radius r cm.
Vertical cross-section, not to scale
(a)
Find the volume of the cup when it is full.
(2)
(b)
Show that r = 4h/9.
(1)
(c)
The cup contains 100 cm3 of water. Find the depth of the water.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (1/3)π(4)2(9)
✓ 1
A1 151 cm3 (accept 48π)
✓ 1
Part (b)
R1 the water forms a cone similar to the cup, so r/h = 4/9
✓ 1
Part (c)
M1 V = (1/3)π(4h/9)2h = 16πh3/243
✓ 1
M1 sets 16πh3/243 = 100
✓ 1
A1 h = 7.85 cm
✓ 1
Exam tip: Express the radius in terms of the depth before using the volume formula, so the equation has only one unknown.
206IBMA2S-048-00-093.1 Three-dimensional geometryPaper 2Medium13 marks
Calculator
A grain silo is modelled as a right circular cylinder of radius 3.2 m and height 12 m. It is topped by a right circular cone of the same radius and vertical height 2.4 m, which forms the roof. The silo stands on a concrete base.
Vertical cross-section, not to scale
(a)
Find the slant height of the conical roof.
(2)
(b)
Find the total volume of the silo.
(3)
(c)
The outside of the silo is painted, apart from the base. Find the area to be painted.
(3)
(d)
Two coats of paint are needed. One litre covers 9.5 m2 and paint is sold in 5-litre cans. Find the number of cans that must be bought.
(3)
(e)
Grain is poured into the silo to a volume of 300 m3, with a flat surface lying within the cylinder. Find the depth of the grain.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 √(3.22 + 2.42)
✓ 1
A1 4.0 m
✓ 1
Part (b)
M1 cylinder π(3.2)2(12)
✓ 1
M1 cone (1/3)π(3.2)2(2.4)
✓ 1
A1 412 m3 (accept 131.072π)
✓ 1
Part (c)
M1 curved surface of cylinder 2π(3.2)(12)
✓ 1
M1 curved surface of cone π(3.2)(4.0)
✓ 1
A1 281 m2 (accept 89.6π)
✓ 1
Part (d)
M1 litres needed: 2 × 281.49… ÷ 9.5 = 59.3 litres
✓ 1
M1 divides by 5: 11.85… cans
✓ 1
A1 rounds up to 12 cans
✓ 1
Part (e)
M1 π(3.2)2d = 300
✓ 1
A1 d = 9.33 m
✓ 1
Exam tip: Read which surfaces are included. Here the base is excluded and the circle where the cone meets the cylinder is hidden inside the silo, so only the two curved surfaces count.
207IBMA2S-048-00-103.1 Three-dimensional geometryPaper 2Medium12 marks
Calculator
A skateboard ramp is a triangular prism. Its horizontal rectangular base ABCD has AB = 3.6 m and BC = 2.5 m. Points E and F lie 1.2 m vertically above B and C respectively. The riding surface is the sloping rectangle ADFE.
Diagram not to scale
(a)
Find the angle between the riding surface and the ground.
(2)
(b)
Find the area of the riding surface.
(3)
(c)
Find the volume of the ramp.
(2)
(d)
A skater rides in a straight line across the riding surface from A to F. Find the distance AF.
(2)
(e)
Find the angle between the line AF and the ground.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 tan θ = 1.2/3.6
✓ 1
A1 θ = 18.4°
✓ 1
Part (b)
M1 AE = √(3.62 + 1.22)
✓ 1
A1 AE = 3.79 m (3.7947…)
✓ 1
A1 area = 3.7947… × 2.5 = 9.49 m2
✓ 1
Part (c)
M1 ½ × 3.6 × 1.2 × 2.5
✓ 1
A1 5.4 m3
✓ 1
Part (d)
M1 √(3.62 + 2.52 + 1.22)
✓ 1
A1 4.54 m
✓ 1
Part (e)
M1 identifies angle FAC, with FC vertical and AC = √(3.62 + 2.52) = 4.383…
✓ 1
M1 sin θ = 1.2/4.544… or tan θ = 1.2/4.383…
✓ 1
A1 θ = 15.3°
✓ 1
Common error: A diagonal path across a slope is less steep than the slope itself. If your angle for AF is larger than 18.4°, the triangle has been set up incorrectly.
208IBMA2S-048-01-023.7 Circular functions and their graphsPaper 2Medium7 marks
Calculator
A test rig for car suspension moves a platform up and down. The height of the platform above the workshop floor, \(h\) cm, at time \(t\) seconds is modelled by \(h(t)=40+6\sin(2.5\pi t)\), \(t\ge 0\).
(a)
Write down the maximum and minimum heights of the platform.
(2)
(b)
Find the period of the motion.
(2)
(c)
Find the number of complete oscillations the platform makes in one minute.
(1)
(d)
Find the first time after \(t=0\) at which the platform is 43 cm above the floor.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 maximum 46 cm
✓ 1
A1 minimum 34 cm
✓ 1
Part (b)
M1 period \(=\frac{2\pi}{2.5\pi}\)
✓ 1
A1 0.8 s
✓ 1
Part (c)
A1 \(\frac{60}{0.8}=75\)
✓ 1
Part (d)
M1 setting \(40+6\sin(2.5\pi t)=43\), either solved on a GDC (intersection or graph) or rearranged to \(\sin(2.5\pi t)=0.5\), giving \(2.5\pi t=\frac{\pi}{6}\)
✓ 1
A1 \(t=\frac{1}{15}\approx 0.0667\) s
✓ 1
Exam tip: For the first time a value is reached, take the principal value of arcsin (or the first intersection on your GDC) and check that it gives a positive t before looking for later solutions.
209IBMA2S-048-01-053.7 Circular functions and their graphsPaper 2Medium7 marks
Calculator
A satellite in low Earth orbit passes over different latitudes as it orbits. Its latitude, \(L\) degrees (positive for north), \(t\) minutes after it crosses the equator heading north, is modelled by \(L(t)=51.6\sin\left(\frac{2\pi t}{92}\right)\), where the argument of the sine function is in radians.
(a)
Write down the time taken for one complete orbit. Hence find the number of complete orbits the satellite makes in one day (1440 minutes).
(2)
(b)
Find the first time at which the satellite reaches latitude 40° north.
(2)
(c)
Find the length of time, in each orbit, for which the satellite is north of latitude 40° north.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 period = 92 minutes
✓ 1
A1 \(\frac{1440}{92}=15.65\ldots\) so 15 complete orbits
✓ 1
Part (b)
M1 \(51.6\sin\left(\frac{2\pi t}{92}\right)=40\), or a graphical method on a GDC
✓ 1
A1 \(t=13.0\) minutes (12.988…)
✓ 1
Part (c)
M1 finding the second time, by symmetry (\(t=46-12.988\)) or from a GDC
✓ 1
M1 \(33.012-12.988\), i.e. subtracting the two times
✓ 1
A1 20.0 minutes
✓ 1
Exam tip: Use the symmetry of a sine curve about its peak. If the first crossing is at t₁ within a half-period of length T/2, the second crossing is at T/2 − t₁.
210IBMA2S-048-01-093.7 Circular functions and their graphsPaper 2Medium12 marks
Calculator
A respiratory nurse models the volume of air, \(V\) litres, in a patient's lungs \(t\) seconds after monitoring begins by \(V(t)=a\sin\big(b(t-c)\big)+d\), where \(a>0\), \(b>0\) and \(0<c<4\). At rest, the patient's lung volume varies between 2.6 litres and 3.4 litres, and the patient takes 15 breaths per minute. The volume first reaches its maximum at \(t=1.5\).
(a)
Find the value of \(a\).
(2)
(b)
Find the value of \(d\).
(1)
(c)
Find the exact value of \(b\).
(2)
(d)
Find the value of \(c\).
(2)
(e)
Find the length of time, during each breath, for which the lung volume exceeds 3.2 litres.
(3)
(f)
After exercise, the patient takes 24 breaths per minute with the same maximum and minimum volumes. Find the new value of \(b\).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{3.4-2.6}{2}\)
✓ 1
A1 \(a=0.4\)
✓ 1
Part (b)
A1 \(d=3\)
✓ 1
Part (c)
M1 period \(=\frac{60}{15}=4\) s
✓ 1
A1 \(b=\frac{2\pi}{4}=\frac{\pi}{2}\)
✓ 1
Part (d)
M1 \(\frac{\pi}{2}(1.5-c)=\frac{\pi}{2}\)
✓ 1
A1 \(c=0.5\)
✓ 1
Part (e)
M1 \(\sin\left(\frac{\pi}{2}(t-0.5)\right)=0.5\)
✓ 1
A1 \(t=\frac{5}{6}\) and \(t=\frac{13}{6}\)
✓ 1
A1 \(\frac{4}{3}=1.33\) s
✓ 1
Part (f)
M1 new period \(=\frac{60}{24}=2.5\) s
✓ 1
A1 \(b=\frac{2\pi}{2.5}=2.51\)
✓ 1
Exam tip: Convert a frequency (breaths per minute) into a period in the units of t (seconds) before using b = 2π / period.
211IBMA2S-048-01-123.7 Circular functions and their graphsPaper 2Medium6 marks
Calculator
The voltage, \(V\) volts, of a household mains electricity supply \(t\) seconds after a meter starts recording is modelled by \(V(t)=325\sin(100\pi t)\).
(a)
Find the period of \(V\).
(1)
(b)
Find the first time at which the voltage reaches 230 V.
(2)
(c)
Find the fraction of each cycle for which the magnitude of the voltage, \(|V|\), is greater than 230 V.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(\frac{2\pi}{100\pi}=0.02\) s
✓ 1
Part (b)
M1 \(325\sin(100\pi t)=230\)
✓ 1
A1 \(t=0.00250\) s (0.0025026…)
✓ 1
Part (c)
M1 positive half-cycle: \(V>230\) between \(t=0.0025026\) and \(t=0.01-0.0025026\), a duration of 0.0049948 s
✓ 1
M1 doubling for the negative half-cycle, where \(V<-230\), and dividing by 0.02
✓ 1
A1 0.499
✓ 1
Exam tip: A condition on |V| is met in both the positive and the negative halves of each cycle. Use symmetry to double the time from one half rather than solving again.
212IBMA2S-048-06-053.5 The unit circle and the ambiguous casePaper 2Medium7 marks
Calculator
A robotic arm has a shoulder joint A, an elbow joint B and a gripper C. The upper segment AB is 40 cm long and the lower segment BC is 30 cm long. For a particular task the controller sets angle BAC = 35°. Let AC = \(x\) cm.
Diagram not to scale
(a)
Show that \(x\) satisfies \(x^2 - (80\cos 35°)x + 700 = 0\).
(2)
(b)
Find the two possible values of \(x\).
(3)
(c)
The controller chooses the larger value of \(x\). Find angle ABC for this position.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 cosine rule with BC opposite angle A: \(30^2 = 40^2 + x^2 - 2(40)x\cos 35°\)
✓ 1
A1 rearranging to \(x^2 - (80\cos 35°)x + 700 = 0\) AG
✓ 1
Part (b)
M1 attempt to solve the quadratic, e.g. \(x^2 - 65.53x + 700 = 0\) or use of the quadratic formula
✓ 1
A1 \(x = 52.1\) cm
✓ 1
A1 \(x = 13.4\) cm
✓ 1
Part (c)
M1 valid method for angle ABC, using either of these routes
• Sine rule: \(\sin C = \frac{40\sin 35°}{30}\) with the acute value \(C = 49.9°\) chosen. The acute value is correct because the larger AC must be opposite the larger angle B, whereas \(C = 130.1°\) gives the smaller triangle with \(x = 13.4\). Then \(B = 180° - 35° - C\).
—
A1 angle ABC = 95.1°
✓ 1
Exam tip: Setting up the cosine rule with the unknown side as \(x\) produces a quadratic, which automatically reveals both triangles in the ambiguous case. When you then use the sine rule, check that the angle you choose matches the triangle you are working in. The cosine rule avoids this ambiguity altogether.
213IBMA2S-048-06-083.3 Applications of trigonometryPaper 2Medium6 marks
Calculator
A mobile phone mast stands vertically with its foot F on level ground. Point A is due west of F and point B is due south of F, and A and B are 150 m apart. The angle of elevation of the top of the mast from A is 28° and from B is 35°.
Diagram not to scale
(a)
Find the height of the mast.
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 expressing AF in terms of height \(h\): \(AF = \frac{h}{\tan 28°}\)
Exam tip: Introduce a letter for the unknown height and write every horizontal distance in terms of it. The right angle at F comes from the compass directions (west and south), not from the mast.
214IBMA2S-048-06-103.3 Applications of trigonometryPaper 2Medium8 marks
Calculator
A tower crane has a vertical mast. A straight jib PQ of length 25 m is pivoted at a point P on the mast, with the end Q above the level of P. A straight cable runs from Q to a point R on the mast, 8 m vertically above P. The length of cable QR can be adjusted to change the angle of the jib.
Diagram not to scale
(a)
When QR = 20 m, find angle QPR.
(3)
(b)
Hence find the height of Q above P.
(2)
(c)
The cable is shortened to 18 m. Find the vertical distance through which the end Q rises.
A1 new height = \(25 \times 0.9125 = 22.8\) m (22.8125)
✓ 1
A1 rise = 22.8125 − 18.0625 = 4.75 m (accept 4.7 m from rounded heights)
✓ 1
Exam tip: Angle QPR is measured from the vertical mast, so the height of Q above P uses cosine, not sine. Keep the full calculator values of the heights and round only the final difference.
215IBMA2S-048-04-053.5 The unit circle and the ambiguous casePaper 2Medium6 marks
Calculator
A long straight fence, more than 100 m in length, runs across a flat field starting from a post A. A second post B is 42 m from A. The fence makes an angle of 28° with the line AB. A goat is tethered to post B by a rope of length 30 m. The rope is not obstructed by the fence. Let x m be the distance along the fence from A to a point where the fully stretched rope just reaches the fence.
A1 rearranges correctly, using 1764 − 900 = 864, to obtain the given equation
✓ 1
Part (b)
M1 solves the quadratic, e.g. using \(84\cos 28° = 74.17\) in the quadratic formula or with a GDC
✓ 1
A1 x = 14.5 or x = 59.7
✓ 1
Part (c)
R1 the points of the fence within 30 m of B lie between the two intersection points
✓ 1
A1 length = 59.69 − 14.47 = 45.2 m
✓ 1
Exam tip: When the given angle is not between the two known sides, the cosine rule produces a quadratic in the unknown side. Two positive roots show directly that two triangles exist, so the ambiguous case of the sine rule is avoided.
216IBMA2S-048-04-063.2 Right-angled and non-right-angled trigonometryPaper 2Medium7 marks
Calculator
A rectangular sports pitch ABCD lies on level ground, with AB = 100 m and AD = 64 m. A vertical floodlight mast AT of height 35 m stands at corner A.
Diagram not to scale
(a)
Find the length of the diagonal AC.
(2)
(b)
Find the angle of elevation of T from C.
(2)
(c)
Find the angle BTD.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(AC^2 = 100^2 + 64^2\)
✓ 1
A1 AC = 119 m (118.7)
✓ 1
Part (b)
M1 \(\tan\theta = \frac{35}{118.7}\)
✓ 1
A1 θ = 16.4°
✓ 1
Part (c)
M1 finds \(TB = \sqrt{11225}\) (105.9 m) and \(TD = \sqrt{5321}\) (72.9 m), using BD = AC
Exam tip: In 3D problems, sketch each triangle you use as a separate flat diagram. Keep squared lengths exact (11225, 5321, 14096) inside the cosine rule to avoid rounding errors.
217IBMA2S-048-04-083.2 Right-angled and non-right-angled trigonometryPaper 2Medium6 marks
Calculator
In a model of a radiotherapy treatment, the skin surface is a straight horizontal line. Two radiation beams enter the skin at points P and Q, which are 9.0 cm apart, and both beams meet at a tumour T below the surface. The beam entering at P makes an angle of 58° with the skin surface, and the beam entering at Q makes an angle of 71° with the skin surface. T lies below the segment PQ.
Diagram not to scale
(a)
Find angle PTQ.
(1)
(b)
Find the distance PT.
(3)
(c)
Find the depth of the tumour below the skin surface.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 angle PTQ = 180° − 58° − 71° = 51°
✓ 1
Part (b)
M1 use of the sine rule with PT opposite the 71° angle
✓ 1
A1 \(PT = \frac{9.0\sin 71°}{\sin 51°}\)
✓ 1
A1 PT = 10.9 cm (10.9499…)
✓ 1
Part (c)
M1 depth = PT sin 58°, using the unrounded PT = 10.9499… (or equivalently depth = QT sin 71° with \(QT = \frac{9.0\sin 58°}{\sin 51°}\) = 9.8212…)
✓ 1
A1 depth = 9.29 cm (9.286…)
✓ 1
Exam tip: In the sine rule, pair each side with the angle opposite it: PT is opposite the angle at Q. Keep the unrounded PT in your calculator for the next step, since using 10.9 gives 9.24 cm instead of 9.29 cm.
218IBMA2S-048-04-093.2 Right-angled and non-right-angled trigonometryPaper 2Medium6 marks
Calculator
A ground station S tracks two navigation satellites, A and B. At a particular instant, SA = 20 200 km, SB = 23 500 km and the angle between the two lines of sight, ASB, is 47°.
M1 cosine rule, e.g. \(\cos A = \frac{20200^2 + 17686^2 - 23500^2}{2(20200)(17686)}\), OR sine rule \(\sin A = \frac{23500\sin 47°}{17686}\)
✓ 1
A1 cos A = 0.236 OR sin A = 0.972
✓ 1
A1 angle SAB = 76.4° (76.35°). Accept 76.3° if it follows from using AB = 17 700 km
✓ 1
Exam tip: Angle SAB is opposite the longest side, SB, so check whether it could be obtuse. The cosine rule settles this directly. With the sine rule, you must rule out 180° − 76.4° = 103.6°. Carry the unrounded value of AB into part (b) so that the final angle is not distorted by rounding.
219IBMA2S-048-04-103.5 The unit circle and the ambiguous casePaper 2Medium13 marks
Calculator
On a golf hole, the hole H is 380 m from the tee T. A golfer's drive travels 240 m in a straight line and lands at B, but its direction deviates by 12° from the line TH. The course is level.
Diagram not to scale
(a)
Find the distance BH.
(3)
(b)
(i) Find angle BHT. (ii) Hence find angle TBH.
(3)
(c)
Find the area of triangle TBH.
(2)
(d)
Explain why the sine rule should not be used directly to find angle TBH.
(2)
(e)
From B, the golfer's second shot travels 150 m in a straight line to C, but its direction deviates by 5° from the line BH. Find the distance CH.
Common error: Rounding BH to 154 before the final part. CH² is a small difference of two large numbers, so rounded intermediate values change the answer noticeably. Keep stored calculator values throughout.
220IBMA3S-3-17r6-063.3 Applications of trigonometryPaper 2Medium7 marks
Calculator
A triangular field ABC has AB = 240 m, AC = 310 m and angle BAC = 64°. Give all answers to 3 significant figures.
Diagram not to scale
(a)
Find the length of the boundary BC.
(3)
(b)
Find the area of the field.
(2)
(c)
Find the shortest distance from B to the side AC.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the cosine rule
✓ 1
A1 \(BC^2=240^2+310^2-2(240)(310)\cos64^\circ\)
✓ 1
A1 BC = 297 m
✓ 1
Part (b)
M1 \(\frac12(240)(310)\sin64^\circ\)
✓ 1
A1 33 400 m2
✓ 1
Part (c)
M1 use area \(=\frac12\times310\times h\) or \(240\sin64^\circ\)
✓ 1
A1 216 m (215.7…)
✓ 1
Exam tip: Use unrounded values from earlier parts in later calculations and round only at the end.
221IBMA3S-3-17r6-073.2 Right-angled and non-right-angled trigonometryPaper 2Medium10 marks
Calculator
A drone hovers at point D, 80 m directly above a point A on level ground. Points B and C are on the ground with AB = 150 m, AC = 210 m and angle BAC = 75°. Give answers to 3 significant figures.
Common error: Angle BDC lies in a vertical plane triangle that is not right-angled, so Pythagoras cannot be used for it.
222IBMA3S-3-17r6-083.7 Circular functions and their graphsPaper 2Medium5 marks
Calculator
The displacement of the tip of a tuning fork is modelled by \(d=2.5\sin(880\pi t)\), where \(d\) is in millimetres and \(t\) is in seconds. Work in radians.
(a)
Find the period of the oscillation in milliseconds, to 3 significant figures.
(2)
(b)
Find the first time, \(t>0\), at which \(d=1.5\) mm. Give your answer in milliseconds to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{2\pi}{880\pi}\)
✓ 1
A1 \(\frac1{440}\) s = 2.27 ms
✓ 1
Part (b)
M1 \(\sin(880\pi t)=0.6\)
✓ 1
M1 \(880\pi t=\sin^{-1}0.6=0.6435\)
✓ 1
A1 \(t=0.233\) ms
✓ 1
Exam tip: The first positive solution comes from the principal value of the inverse sine; the second is \(\pi\) minus it.
223IBMA3S-3-17r6-103.4 Radian measure, arcs and sectorsPaper 2Medium6 marks
Calculator
A circular irrigation field has centre O and radius 120 m. A straight fence AB joins two points on its edge and angle AOB = 1.9 radians. The region between the fence and the arc AB is a segment. Give answers to 1 decimal place unless stated.
Diagram not to scale
(a)
Find the length of the fence AB.
(2)
(b)
Find the area of the segment, to the nearest m2.
(3)
(c)
Find the perimeter of the segment.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(2(120)\sin0.95\) or cosine rule
✓ 1
A1 195.2 m
✓ 1
Part (b)
M1 sector area \(\frac12(120)^2(1.9)\)
✓ 1
M1 subtract triangle area \(\frac12(120)^2\sin1.9\)
✓ 1
A1 6867 m2 (6866.6…)
✓ 1
Part (c)
A1 \(120(1.9)+195.2=423.2\) m
✓ 1
Common error: The segment is sector minus triangle; adding them gives a different region.
224IBMA3S-3-17r6-113.8 Trigonometric equationsPaper 2Medium6 marks
Calculator
Solve \(3\sin2x=2\cos x\) for \(0^\circ\le x\le360^\circ\). Give non-exact answers to 1 decimal place.
(a)
Solve the equation.
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use \(\sin2x=2\sin x\cos x\)
✓ 1
M1 collect and factorise: \(2\cos x(3\sin x-1)=0\)
✓ 1
A1 \(\cos x=0\) or \(\sin x=\frac13\)
✓ 1
A1 \(x=90^\circ\), \(270^\circ\)
✓ 1
A1 \(x=19.5^\circ\)
✓ 1
A1 \(x=160.5^\circ\) (160.53…)
✓ 1
Common error: Never divide both sides by cos x; doing so loses the solutions where cos x = 0.
225IBMA3S-3-17r8-053.5 The unit circle and the ambiguous casePaper 2Medium5 marks
Calculator
Two spotlights are fixed at points A and B on a lighting bar, with AB = 9.4 m. A performer stands at point P on the stage with AP = 12.6 m and BP = 7.8 m.
Diagram not to scale
(a)
Find angle APB, to the nearest tenth of a degree.
(3)
(b)
Find the area of triangle APB, to 3 significant figures. Use the unrounded angle from (a).
Exam tip: When the three sides are known, use the cosine rule with the angle you want opposite the side that appears alone on the left.
226IBMA3S-3-17r8-063.5 The unit circle and the ambiguous casePaper 2Medium8 marks
Calculator
Two ships leave harbour H at the same time. Ship A sails 6 km on a bearing of 040°. Ship B sails 9 km on a bearing of 125°.
Diagram not to scale
(a)
Find the size of angle AHB.
(1)
(b)
Find the distance AB, to 3 significant figures.
(3)
(c)
Find the bearing of B from A, to 1 decimal place. Use unrounded values from (b).
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(125^\circ-40^\circ=85^\circ\)
✓ 1
Part (b)
M1 use of the cosine rule \(AB^2=6^2+9^2-2(6)(9)\cos85^\circ\)
✓ 1
A1 \(AB^2=107.58\ldots\)
✓ 1
A1 \(AB=10.4\) km
✓ 1
Part (c)
M1 attempt to use the sine rule \(\frac{\sin \angle HAB}{9}=\frac{\sin85^\circ}{AB}\)
✓ 1
A1 \(\angle HAB=59.8^\circ\)
✓ 1
M1 bearing of H from A is \(220^\circ\), subtract \(59.8\ldots^\circ\)
✓ 1
A1 \(160.2^\circ\)
✓ 1
Common error: The bearing of H from A is the bearing of A from H plus 180°, not the same number; use a sketch with a north line at each point.
227IBMA3S-3-17r8-083.3 Applications of trigonometryPaper 2Medium6 marks
Calculator
A lighthouse L sweeps a beam over a sector of sea of radius 14 km. The angle at L between the two edges of the sector is 75°. The two edges meet the sea boundary at points P and Q.
Diagram not to scale
(a)
Find the length of arc PQ, to 3 significant figures.
(2)
(b)
The straight line PQ cuts the sector into a triangle LPQ and a segment. Find the area of the segment, to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 convert \(75^\circ=\frac{5\pi}{12}\) rad and use \(s=r\theta\)
✓ 1
A1 \(18.3\) km
✓ 1
Part (b)
M1 area of sector \(\frac12(14)^2\frac{5\pi}{12}=128.28\ldots\)
✓ 1
M1 area of triangle \(\frac12(14)^2\sin75^\circ=94.66\ldots\)
✓ 1
M1 subtract sector minus triangle
✓ 1
A1 \(33.6\) km\(^2\)
✓ 1
Exam tip: A segment is a sector minus a triangle; write both areas separately before subtracting.
228IBMA3S-3-17r8-093.8 Trigonometric equationsPaper 2Medium5 marks
Calculator
A theatre spotlight swings so that the angle \(x\) (in radians) between its beam and the vertical satisfies \(\tan 2x=1.5\), where \(0\le x\le\pi\).
(a)
Find all solutions for \(x\), to 3 significant figures.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 change of variable: \(0\le 2x\le 2\pi\)
✓ 1
M1 \(2x=\tan^{-1}1.5=0.9828\ldots\)
✓ 1
A1 \(x=0.491\)
✓ 1
M1 second solution \(2x=0.9828\ldots+\pi\)
✓ 1
A1 \(x=2.06\)
✓ 1
Common error: After doubling the angle, the interval for \(2x\) is twice as long, so a tangent equation gives two solutions for \(x\) here, not one.
229IBMA3S-3-17r8-103.1 Three-dimensional geometryPaper 2Medium7 marks
Calculator
A model of a volcano is a right circular cone with base radius 2.4 km and vertical height 0.75 km.
Diagram not to scale
(a)
Find the angle between the slope of the volcano and the horizontal, to 3 significant figures.
(2)
(b)
Find the volume of the model, to 3 significant figures.
(2)
(c)
Find the curved surface area of the model, to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\tan\alpha=\frac{0.75}{2.4}\)
✓ 1
A1 \(17.4^\circ\)
✓ 1
Part (b)
M1 use of \(V=\frac13\pi r^2h\)
✓ 1
A1 \(\frac13\pi(2.4)^2(0.75)=4.52\) km\(^3\)
✓ 1
Part (c)
M1 slant height \(\sqrt{2.4^2+0.75^2}=2.51\ldots\) km
✓ 1
A1 use of \(\pi r l\)
✓ 1
A1 \(19.0\) km\(^2\)
✓ 1
Common error: The curved surface area uses the slant height \(l\), not the vertical height; find \(l\) with Pythagoras first.
230IBMA3S-3-17r8-113.3 Applications of trigonometryPaper 2Medium5 marks
Calculator
The top of a lighthouse is 42 m above sea level. A sailor looks down at a ship from the top of the lighthouse. Treat the sea as level and the ship as at sea level.
Diagram not to scale
(a)
The angle of depression of the ship is 3.5°. Find its horizontal distance from the lighthouse, to the nearest metre.
(2)
(b)
The ship sails directly towards the lighthouse until the angle of depression is 6.0°. Find how far the ship has sailed, to the nearest metre. Use unrounded values.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\tan3.5^\circ=\frac{42}{d}\)
✓ 1
A1 \(d=687\) m
✓ 1
Part (b)
M1 new distance \(\frac{42}{\tan6^\circ}=399.60\ldots\)
✓ 1
M1 subtract from \(686.69\ldots\)
✓ 1
A1 \(287\) m
✓ 1
Common error: The angle of depression equals the angle of elevation from the ship (alternate angles); it is measured from the horizontal, not from the lighthouse wall.
231IBMA3S-3-16r18-053.4 Radian measure, arcs and sectorsPaper 2Medium7 marks
Calculator
A decorative window in a library is part of a circle with centre O and radius 2.5 m. A chord AB subtends an angle of 1.6 radians at O. The shaded region lies between the chord AB and the minor arc AB.
Diagram not to scale
(a)
Find the length of the arc AB.
(2)
(b)
Find the area of the shaded region, giving your answer to 3 significant figures.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(s=r\theta=2.5\times1.6\)
✓ 1
A1 \(s=4\) m
✓ 1
Part (b)
M1 area of sector \(=\frac{1}{2}(2.5)^2(1.6)\)
✓ 1
M1 area of triangle \(OAB=\frac{1}{2}(2.5)^2\sin1.6\)
✓ 1
A1 triangle area \(=3.12\ldots\) m2
✓ 1
M1 subtract triangle from sector (sector area 5)
✓ 1
A1 area \(=1.88\) m2
✓ 1
Exam tip: Set the calculator to radian mode when a trigonometric function has an angle in radians.
232IBMA3S-3-16r18-073.7 Circular functions and their graphsPaper 2Medium7 marks
Calculator
The wind speed \(v\) m s−1 at a wind farm, \(t\) hours after midnight, is modelled by \(v(t)=7+4\cos\left(\frac{\pi t}{12}\right)\) for \(0\le t\le24\).
(a)
Write down the maximum and minimum wind speeds predicted by the model.
(2)
(b)
Find the period of the function.
(1)
(c)
Find the values of \(t\) for which \(v=9\).
(3)
(d)
The turbines only generate full power when the wind speed is at least 9 m s−1. Find the total time in one day for which this happens.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 maximum \(=11\) m s−1
✓ 1
A1 minimum \(=3\) m s−1
✓ 1
Part (b)
A1 \(\frac{2\pi}{\pi/12}=24\) hours
✓ 1
Part (c)
M1 \(4\cos\left(\frac{\pi t}{12}\right)=2\), so \(\cos\left(\frac{\pi t}{12}\right)=\frac{1}{2}\)
✓ 1
A1 \(t=4\)
✓ 1
A1 \(t=20\)
✓ 1
Part (d)
A1 \(4+(24-20)=8\) hours (ECF from (c))
✓ 1
Exam tip: After finding the first solution of a cosine equation, use the symmetry of the graph over one period to find the second.
233IBMA3S-3-16r18-133.5 The unit circle and the ambiguous casePaper 2Medium7 marks
Calculator
In triangle ABC, which models a rope course in a swimming centre, \(AB=7.5\) m, \(BC=5.2\) m and angle \(BAC=40^\circ\). Angle \(ACB\) can take two different values.
Diagram not to scale
(a)
Find the two possible sizes of angle \(ACB\), giving your answers to 3 significant figures.
(4)
(b)
Find the smaller of the two possible lengths of \(AC\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sine rule \(\frac{\sin ACB}{7.5}=\frac{\sin40^\circ}{5.2}\)
✓ 1
A1 \(\sin ACB=0.9271\ldots\)
✓ 1
A1 \(ACB=68.0^\circ\)
✓ 1
A1 \(ACB=112^\circ\) (180° minus the first value)
✓ 1
Part (b)
M1 uses the obtuse case \(ACB=112.01\ldots^\circ\) so \(ABC=180-40-112.01\ldots=27.98\ldots^\circ\)
✓ 1
M1 sine rule \(AC=\frac{5.2\sin ABC}{\sin40^\circ}\)
✓ 1
A1 \(AC=3.80\) m
✓ 1
Common error: When the sine rule is used to find an angle, check whether \(180^\circ\) minus the angle is also possible before rejecting it.
234IBMA3S-3-17r24-053.3 Applications of trigonometryPaper 2Medium7 marks
Calculator
A volcano has its summit at S. Two observers, at P and Q, stand on level ground in a straight line with the point F directly below the summit, with P between Q and F. The distance PQ is 1500 m. The angle of elevation of S is 21° from P and 14° from Q.
Diagram not to scale
(a)
Show that angle PSQ = 7°.
(2)
(b)
Find the distance PS, to the nearest metre. Keep the unrounded value for part (c).
(3)
(c)
Find the height of the summit above the ground, to the nearest metre.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 identifies angle SPF = 21° as the exterior angle of triangle SPQ (or angle SPQ = 159°)
M1 sine rule in triangle PQS: PS/sin 14° = 1500/sin 7°
✓ 1
A1 correct substitution and rearrangement
✓ 1
A1 PS = 2977.6 = 2978 m
✓ 1
Part (c)
M1 height = PS sin 21°
✓ 1
A1 1067.1 = 1067 m
✓ 1
Common error: The angle at the summit is the difference of the two elevation angles, not their sum; draw the vertical triangle and mark the exterior angle.
235IBMA3S-3-17r24-063.2 Right-angled and non-right-angled trigonometryPaper 2Medium7 marks
Calculator
A triangular platform ABC is built on a theatre stage. AB = 7.2 m, AC = 9.5 m and angle BAC = 64°. Give answers to 3 significant figures, using unrounded values in later parts.
Exam tip: Use the two sides and the included angle for the area formula; if you then need a height, rearrange the area instead of starting a new triangle.
236IBMA3S-3-17r24-073.3 Applications of trigonometryPaper 2Medium7 marks
Calculator
Ship B is 15 km from port P on a bearing of 062°. Ship C is 22 km from P on a bearing of 140°.
Diagram not to scale
(a)
Show that angle BPC = 78°.
(1)
(b)
Find the distance BC, to 3 significant figures.
(3)
(c)
Find the bearing of C from B, to the nearest degree.
M1 sine rule to find angle PCB: sin C/15 = sin 78°/BC, so angle PCB = 37.8°
✓ 1
M1 angle PBC = 180° − 78° − 37.8° = 64.2° and the bearing of P from B is 242°
✓ 1
A1 242° − 64.2° = 177.8° ≈ 178°
✓ 1
Common error: A bearing is measured clockwise from north at the starting point; to go back from B to P add 180° to the original bearing before using the angle in the triangle.
237IBMA3S-3-17r24-083.4 Radian measure, arcs and sectorsPaper 2Medium7 marks
Calculator
A circular revolving stage has radius 6.4 m and centre O. A curtain is hung along a chord AB of the stage edge, where angle AOB = 1.85 radians. The region between the chord and the minor arc AB is a segment. Give answers to 3 significant figures; the calculator must be in radian mode.
Diagram not to scale
(a)
Find the length of the minor arc AB.
(2)
(b)
Find the length of the curtain AB.
(2)
(c)
Find the area of the segment.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 l = rθ = 6.4 × 1.85
✓ 1
A1 11.84 = 11.8 m
✓ 1
Part (b)
M1 cosine rule or 2 × 6.4 × sin(1.85/2)
✓ 1
A1 10.22 ≈ 10.2 m
✓ 1
Part (c)
M1 area of sector ½(6.4)2(1.85) = 37.888 m2
✓ 1
M1 area of triangle ½(6.4)2 sin 1.85 = 19.687... m2 and subtract
✓ 1
A1 18.201 ≈ 18.2 m2
✓ 1
Exam tip: Segment area is the sector area minus the triangle area; the triangle needs the sine of the angle, so check your calculator is in radians.
238IBMA3S-3-17r24-093.5 The unit circle and the ambiguous casePaper 2Medium7 marks
Calculator
A ship's route is modelled by triangle PQR with PQ = 9.0 km, QR = 7.0 km and angle QPR = 48°. Two different triangles satisfy these conditions.
(a)
Find the two possible sizes of angle PRQ, to 1 decimal place.
(3)
(b)
Find the two possible lengths of PR, to 3 significant figures. Use unrounded angles.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sine rule: sin R/9.0 = sin 48°/7.0
✓ 1
A1 R = 72.8°
✓ 1
A1 R = 180° − 72.8° = 107.2°
✓ 1
Part (b)
M1 angle PQR = 180° − 48° − R (59.2° or 24.8°)
✓ 1
M1 sine rule PR = 7.0 sin Q/sin 48°
✓ 1
A1 PR = 8.09 ≈ 8.09 km
✓ 1
A1 PR = 3.957 ≈ 3.96 km
✓ 1
Common error: When sin R gives an acute angle, always test 180° minus that angle as well; both are valid only if the angle sum with the given angle is below 180°.
239IBMA3S-3-17r24-113.1 Three-dimensional geometryPaper 2Medium6 marks
Calculator
A theatre prop is a solid right pyramid with a square base of side 1.2 m. The apex is vertically above the centre of the base, at a height of 0.9 m. Give angles to 3 significant figures.
Diagram not to scale
(a)
Find the distance from the centre of the base to one corner of the base.
(2)
(b)
Find the angle between a sloping edge and the base.
(2)
(c)
Find the angle between a triangular face and the base.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 diagonal = 1.2√2 (Pythagoras) and half of it
✓ 1
A1 0.8485 ≈ 0.849 m
✓ 1
Part (b)
M1 tan θ = 0.9 / (distance from (a))
✓ 1
A1 θ = 46.69° ≈ 46.7°
✓ 1
Part (c)
M1 tan φ = 0.9/0.6, using the distance from the centre to the midpoint of a base edge
✓ 1
A1 φ = 56.31° ≈ 56.3°
✓ 1
Common error: The angle between a face and the base is measured to the midpoint of a base edge (perpendicular to the edge), not to a corner.
240IBMA3S-3-17r25-063.7 Circular functions and their graphsPaper 2Medium10 marks
Calculator
The ride height of a Formula 1 car's floor above the track as it crosses a bumpy kerb is modelled by h(t) = a sin(bt) + d, for t ≥ 0, where h is in millimetres, t is in seconds and bt is in radians. The ride height has a maximum of 44 mm and a minimum of 32 mm. At t = 0 the height is at its mean value and is increasing. The time between successive maximum heights is 0.25 s.
(a)
Find the value of a and the value of d.
(2)
(b)
Find the value of b, giving your answer in terms of π.
(2)
(c)
Find the first time, t > 0, at which the ride height is 41 mm. Give your answer to 3 significant figures.
(3)
(d)
Find the proportion of each cycle for which the ride height is greater than 41 mm.
M1 the height exceeds 41 mm while \(\frac{\pi}{6}\) < 8πt < \(\frac{5\pi}{6}\)
✓ 1
A1 this is an interval of \(\frac{2\pi}{3}\) in a full cycle of 2π
✓ 1
A1 proportion = \(\frac{1}{3}\)
✓ 1
Exam tip: Maximum and minimum values give the amplitude and the vertical shift: amplitude = (max − min) ÷ 2 and shift = (max + min) ÷ 2.
241IBMA3S-3-17r25-093.3 Applications of trigonometryPaper 2Medium8 marks
Calculator
A medical helicopter leaves its base H and flies 24 km on a bearing of 072° to the scene of an accident at A. It then flies 31 km on a bearing of 155° to a hospital at B.
Diagram not to scale
(a)
Show that angle HAB = 97°.
(2)
(b)
Calculate the direct distance HB, giving your answer to 3 significant figures.
(3)
(c)
Find the bearing of B from H, giving your answer to the nearest degree. Use the unrounded value of HB.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 the bearing of H from A is 072° + 180° = 252°
✓ 1
A1 angle HAB = 252° − 155° = 97° (AG)
✓ 1
Part (b)
M1 attempt to use the cosine rule
✓ 1
A1 HB2 = 242 + 312 − 2(24)(31)cos97°
✓ 1
A1 HB = 41.5 km
✓ 1
Part (c)
M1 use the sine rule: \(\frac{\sin AHB}{31} = \frac{\sin 97°}{HB}\)
✓ 1
A1 angle AHB = 47.9°
✓ 1
A1 bearing = 072° + 47.9° = 120°
✓ 1
Exam tip: Draw a sketch with a north line at each point; the bearing of the return journey is the outward bearing plus or minus 180°.
242IBMA3S-3-19r27-053.5 The unit circle and the ambiguous casePaper 2Medium7 marks
Calculator
Three mobile-phone masts A, B and C stand on level ground. AB = 8.4 km, AC = 6.1 km and angle BAC = 62°.
Diagram not to scale
(a)
Find BC, in km to 3 significant figures.
(2)
(b)
Find the size of angle ABC, to the nearest 0.1°. Use the unrounded value of BC.
(2)
(c)
Find the area of triangle ABC, in km2 to 3 significant figures.
(2)
(d)
Explain why angle ABC cannot be obtuse.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(BC^2=8.4^2+6.1^2-2(8.4)(6.1)\cos62°\)
✓ 1
A1 BC = 7.72 km
✓ 1
Part (b)
M1 \(\frac{\sin B}{6.1}=\frac{\sin62°}{BC}\)
✓ 1
A1 angle ABC = 44.2°
✓ 1
Part (c)
M1 \(\frac12\times8.4\times6.1\times\sin62°\)
✓ 1
A1 22.6 km2
✓ 1
Part (d)
R1 AC is shorter than BC, so the angle opposite AC is smaller than angle BAC = 62°, hence acute
✓ 1
Common error: When the sine rule is used to find an angle, check whether an obtuse angle is possible: the angle opposite the shorter side is the smaller one.
243IBMA3S-3-19r27-073.7 Circular functions and their graphsPaper 2Medium8 marks
Calculator
The temperature \(T\) °C at a mountain camp, \(t\) hours after midnight, is modelled by \(T=-6+8\sin\left(\frac{\pi}{12}(t-9)\right)\) for \(0\le t\le24\), where the angle is in radians.
(a)
Write down the maximum temperature and the minimum temperature predicted by the model.
(2)
(b)
Find the two times at which the temperature is exactly 0 °C. Give each time in hours to 1 decimal place.
(4)
(c)
Hence find the length of time, in hours to 3 significant figures, for which the temperature is above 0 °C. Use the unrounded times from (b).
M1 second angle \(\pi-0.848\ldots\) (symmetry of the sine curve)
✓ 1
A1 second time \(t=17.8\)
✓ 1
Part (c)
M1 difference of the two unrounded times, 17.7606 − 12.2394
✓ 1
A1 5.52 hours (ECF from (b))
✓ 1
Common error: Stopping after the calculator's first answer to \(\sin\theta=0.75\) loses the second solution in the same cycle, \(\pi-\theta\).
244IBMA3S-3-19r27-083.1 Three-dimensional geometryPaper 2Medium7 marks
Calculator
A tent is a right pyramid. Its base ABCD is a horizontal square of side 3.6 m with centre O. The apex V is vertically above O and VO = 2.4 m. The point M is the midpoint of BC.
Diagram not to scale
(a)
Find the length of the diagonal AC, in m to 3 significant figures.
(2)
(b)
Find the angle between the edge VA and the base, to the nearest 0.1°.
(2)
(c)
Find the angle between VM and the base, to the nearest 0.1°.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(AC=\sqrt{3.6^2+3.6^2}\)
✓ 1
A1 5.09 m
✓ 1
Part (b)
M1 \(\tan\theta=\dfrac{2.4}{AO}\) with \(AO=\frac12AC=2.5456\ldots\)
✓ 1
A1 43.3°
✓ 1
Part (c)
M1 \(OM=\frac12\times3.6=1.8\) m
✓ 1
M1 \(\tan\phi=\dfrac{2.4}{1.8}\)
✓ 1
A1 53.1°
✓ 1
Exam tip: For an angle between a line and a plane, drop the perpendicular to the plane and use the right-angled triangle formed.
245IBMA3S-3-19r27-113.8 Trigonometric equationsPaper 2Medium8 marks
Calculator
The voltage \(V\) volts across a component in a relay station is modelled by \(V=1+4\sin t\cos t\) for \(0\le t\le\pi\), where \(t\) is in milliseconds and the angle is in radians.
(a)
Show that \(V=1+2\sin2t\).
(2)
(b)
Write down the maximum and minimum values of \(V\).
(2)
(c)
Find the two times at which \(V=2.5\). Give each answer to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of \(\sin2t=2\sin t\cos t\)
✓ 1
A1 \(4\sin t\cos t=2\sin2t\), so \(V=1+2\sin2t\) (AG)
✓ 1
Part (b)
A1 maximum 3
✓ 1
A1 minimum −1
✓ 1
Part (c)
M1 \(\sin2t=0.75\)
✓ 1
A1 \(t=0.424\)
✓ 1
M1 second solution from \(2t=\pi-0.848\ldots\)
✓ 1
A1 \(t=1.15\)
✓ 1
Common error: Because the interval for \(t\) is \(0\le t\le\pi\), the angle \(2t\) runs over a full turn, so \(\sin2t=0.75\) has exactly two solutions.
246IBMA3S-3-19r30-053.3 Applications of trigonometryPaper 2Medium6 marks
Calculator
Two climbers leave a base camp C at the same time. Climber P walks 4.2 km on a bearing of 040°. Climber Q walks 5.1 km on a bearing of 115°.
Diagram not to scale
(a)
Show that angle PCQ = 75°.
(1)
(b)
Find the distance PQ, giving your answer to 3 significant figures.
(3)
(c)
Find the area of triangle CPQ, giving your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 115° − 40° = 75°
✓ 1
Part (b)
M1 use of the cosine rule
✓ 1
A1 PQ² = 4.2² + 5.1² − 2(4.2)(5.1)cos 75°
✓ 1
A1 PQ = 5.71 km
✓ 1
Part (c)
M1 area = ½(4.2)(5.1) sin 75°
✓ 1
A1 10.3 km²
✓ 1
Exam tip: Bearings are measured clockwise from north, so the angle between two bearings is found by subtraction.
247IBMA3S-3-19r30-063.5 The unit circle and the ambiguous casePaper 2Medium7 marks
Calculator
Two radio masts P and Q are 9.5 km apart. A relay station R is 7.0 km from Q, and angle QPR = 42°. Use unrounded values in later working and give angles to 1 decimal place and lengths to 3 significant figures.
Diagram not to scale
(a)
Find the two possible values of angle PRQ.
(4)
(b)
Find the two possible values of the distance PR.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of the sine rule: sin R ⁄ 9.5 = sin 42° ⁄ 7.0
✓ 1
A1 sin R = 0.908…
✓ 1
A1 65.2°
✓ 1
A1 114.8°
✓ 1
Part (b)
M1 angle PQR = 180° − 42° − R and use of the sine rule
✓ 1
A1 9.99 km
✓ 1
A1 4.13 km
✓ 1
Common error: When the sine rule is used to find an angle, always check whether 180° minus the calculator value also fits the triangle.
248IBMA3S-3-19r30-083.1 Three-dimensional geometryPaper 2Medium7 marks
Calculator
A chocolate is made in the shape of a right pyramid VABCD. The base ABCD is a horizontal square of side 6 cm and the apex V is 4 cm vertically above the centre of the base. Give angles to 1 decimal place.
Diagram not to scale
(a)
Show that the length of edge VA is √34 cm, which is 5.83 cm to 3 significant figures.
(3)
(b)
Find the angle between edge VA and the base.
(2)
(c)
Find the angle between face VAB and the base.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 half the diagonal of the base = ½ × 6√2 = 3√2
✓ 1
M1 VA² = (3√2)² + 4² = 18 + 16
✓ 1
A1 VA = √34 = 5.83 cm (AG)
✓ 1
Part (b)
M1 tan(angle) = 4 ⁄ (3√2)
✓ 1
A1 43.3°
✓ 1
Part (c)
M1 the horizontal distance from the centre to the midpoint of AB is 3, so tan(angle) = 4⁄3
✓ 1
A1 53.1°
✓ 1
Common error: The angle between a sloping face and the base is measured to the midpoint of a base edge, not to a corner of the base.
249IBMA3S-3-19r30-093.4 Radian measure, arcs and sectorsPaper 2Medium10 marks
Calculator
A circular mountain lake has centre O and radius 220 m. Points A and B lie on the shore, with straight-line distance AB = 300 m. Angle AOB is θ radians and the minor arc AB follows the shore. Use the unrounded value of θ in later parts and give lengths and areas to 3 significant figures.
Diagram not to scale
(a)
Find θ to 3 significant figures.
(3)
(b)
Find the length of the minor arc AB.
(2)
(c)
Find the area of the region between the chord AB and the minor arc AB.
(3)
(d)
A walker goes from A to B along the shore and returns along the straight chord. Find the total distance walked.
M1 correct substitution of r = 220 and θ = 1.5004…
✓ 1
A1 12 200 m²
✓ 1
Part (d)
M1 arc length + chord length
✓ 1
A1 630 m
✓ 1
Exam tip: Keep the unrounded angle in your calculator memory; using 1.50 instead can change the last digit of a later answer.
250IBMA3S-3-19r30-113.5 The unit circle and the ambiguous casePaper 2Medium6 marks
Calculator
A triangular glacier field ABC has AB = 1.8 km and AC = 2.6 km. Angle BAC is acute and the area of the field is 2.1 km². Use the unrounded angle in part (b).
Diagram not to scale
(a)
Find angle BAC, to 3 significant figures.
(3)
(b)
Find the length of BC, to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 ½(1.8)(2.6) sin A = 2.1
✓ 1
A1 sin A = 0.897…
✓ 1
A1 A = 63.8°
✓ 1
Part (b)
M1 use of the cosine rule
✓ 1
A1 BC² = 1.8² + 2.6² − 2(1.8)(2.6) cos A
✓ 1
A1 BC = 2.42 km
✓ 1
Common error: Because A is stated to be acute, take the calculator value of inverse sine; do not also use 180° minus that value.
251IBMA3S-3-19r31-043.2 Right-angled and non-right-angled trigonometryPaper 2Medium7 marks
Calculator
A triangular steel brace PQR on a footbridge has PQ = 14.2 m and PR = 18.6 m, with angle QPR = 71°.
Diagram not to scale
(a)
Find the length QR, correct to 3 significant figures.
(3)
(b)
Find the area of triangle PQR, correct to 3 significant figures.
(2)
(c)
Find the size of the smallest angle in the triangle, correct to the nearest 0.1°. Use the unrounded value of QR.
M1 smallest angle is opposite the shortest side PQ; sine rule sin R/14.2 = sin 71°/QR
✓ 1
A1 angle PRQ = 43.8°
✓ 1
Exam tip: The smallest angle is always opposite the shortest side, and it is always acute, so the sine rule gives no ambiguity there.
252IBMA3S-3-19r31-053.5 The unit circle and the ambiguous casePaper 2Medium7 marks
Calculator
A rail through point A carries a sliding bracket B on a spacecraft boom. A rigid boom AC has length 11.0 m, and a cable BC of length 8.0 m joins C to the bracket B. Angle CAB = 35°. Triangle ABC is drawn from these data.
Diagram not to scale
(a)
Find the two possible sizes of angle ABC, correct to the nearest 0.1°.
(4)
(b)
Find the larger of the two possible lengths of AB, correct to 3 significant figures. Use unrounded values.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sine rule: sin B/11.0 = sin 35°/8.0
✓ 1
A1 B = 52.1°
✓ 1
M1 the other possibility is 180° − 52.1°
✓ 1
A1 B = 127.9°
✓ 1
Part (b)
M1 angle ACB = 180° − 35° − 52.06° = 92.94°
✓ 1
M1 AB/sin ACB = 8.0/sin 35°
✓ 1
A1 AB = 13.9 m
✓ 1
Common error: When you are given two sides and an angle that is not between them, always check whether 180° minus the calculator angle also fits the triangle.
253IBMA3S-3-19r31-063.7 Circular functions and their graphsPaper 2Medium7 marks
Calculator
The temperature \(T\) °C inside a roasting drum is modelled by \(T=d+a\sin(bt)\), where \(t\) is the time in seconds after a reading is started, \(a>0\) and \(b>0\). The temperature oscillates between a minimum of 165 °C and a maximum of 195 °C, and one complete cycle takes 40 seconds.
(a)
Find the values of \(a\), \(b\) and \(d\). Give \(b\) as an exact value.
(3)
(b)
Find the first time at which \(T=188\), correct to 3 significant figures.
(2)
(c)
In each cycle the temperature is above 188 °C for a period of time. Find this time to 3 significant figures, using the unrounded value of the answer to part (b).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(a=15\)
✓ 1
A1 \(d=180\)
✓ 1
A1 \(b=\frac{2\pi}{40}=\frac{\pi}{20}\)
✓ 1
Part (b)
M1 \(15\sin\left(\frac{\pi t}{20}\right)=8\)
✓ 1
A1 \(t=3.58\) s
✓ 1
Part (c)
M1 uses symmetry: duration = 20 − 2t₁
✓ 1
A1 12.8 s
✓ 1
Exam tip: Sketching one cycle of the sine curve and marking the two crossing points shows quickly why the second solution is 20 − t₁.
254IBMA3S-3-19r31-093.4 Radian measure, arcs and sectorsPaper 2Medium9 marks
Calculator
The arch of a footbridge is a circular segment. The arch has a horizontal chord (the span) of 32 m and lies on a circle of radius 20 m. The arch curves upward from the chord. Let θ be the angle, in radians, that the chord subtends at the centre of the circle.
Diagram not to scale
(a)
Find \(\theta\), correct to 3 significant figures.
(2)
(b)
Find the length of the curved edge of the arch, correct to 3 significant figures. Use the unrounded value of θ.
(2)
(c)
Find the greatest height of the arch above the chord.
(2)
(d)
Find the area of the face of the arch between the curve and the chord, correct to 3 significant figures. Use the unrounded value of θ.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\sin\frac{\theta}{2}=\frac{16}{20}\)
✓ 1
A1 θ = 1.85 radians
✓ 1
Part (b)
M1 arc length = 20θ
✓ 1
A1 37.1 m
✓ 1
Part (c)
M1 centre-to-chord distance = √(20² − 16²) = 12
✓ 1
A1 height = 20 − 12 = 8 m
✓ 1
Part (d)
M1 area = sector − triangle
✓ 1
M1 \(\frac{1}{2}(20^2)(\theta-\sin\theta)\)
✓ 1
A1 179 m²
✓ 1
Exam tip: The segment area is the sector minus the triangle formed by the two radii and the chord; draw the radii to see it.
255IBMA3S-3-19r31-113.1 Three-dimensional geometryPaper 2Medium10 marks
Calculator
A bridge pylon is a solid pyramid with a square base ABCD of side 8 m. The apex V is 15 m vertically above the centre of the base.
Diagram not to scale
(a)
Find the length of the sloping edge VA, correct to 3 significant figures.
(3)
(b)
Find the angle between VA and the base, correct to 3 significant figures.
(2)
(c)
Find the angle between a triangular face and the base, correct to 3 significant figures.
(2)
(d)
Find the total area of the four triangular faces, correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 half the diagonal of the base = 4√2
✓ 1
M1 VA² = (4√2)² + 15²
✓ 1
A1 VA = 16.0 m
✓ 1
Part (b)
M1 tan θ = 15/(4√2)
✓ 1
A1 69.3°
✓ 1
Part (c)
M1 tan φ = 15/4 using the midpoint of a base edge
✓ 1
A1 75.1°
✓ 1
Part (d)
M1 slant height = √(15² + 4²)
✓ 1
M1 area = 4 × ½ × 8 × slant height
✓ 1
A1 248 m²
✓ 1
Exam tip: In pyramid questions, draw the right-angled triangle that contains the height and label which horizontal distance it uses: half a side for a face, half a diagonal for an edge.
256IBMA3S-3-18r40-043.5 The unit circle and the ambiguous casePaper 2Medium6 marks
Calculator
Mission control tracks a space probe \(P\) and a relay satellite \(R\) from a ground station \(G\). Distances are in thousands of kilometres. In triangle \(GPR\), \(GR = 8.4\), \(PR = 5.5\) and angle \(PGR = 32^\circ\). The position of \(P\) is not yet known exactly, so there are two possible triangles.
Diagram not to scale
(a)
Find the two possible values of angle \(GPR\). Give your answers to 3 significant figures.
(3)
(b)
Find the two possible values of \(GP\). Give your answers to 3 significant figures, using unrounded angles.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sine rule: \(\frac{\sin GPR}{8.4} = \frac{\sin 32^\circ}{5.5}\)
Exam tip: When the given angle is opposite the shorter of the two known sides, the sine rule can give two triangles: always check \(180^\circ - \theta\) as well.
257IBMA3S-3-18r40-063.3 Applications of trigonometryPaper 2Medium7 marks
Calculator
A drone surveys the site of a new bridge. It starts at point \(A\), flies \(8\) km on a bearing of \(065^\circ\) to point \(B\), returns to \(A\), then flies \(11\) km on a bearing of \(140^\circ\) to point \(C\).
Diagram not to scale
(a)
Find the distance \(BC\). Give your answer to 3 significant figures.
(3)
(b)
Find the bearing of \(C\) from \(B\), using the unrounded value of \(BC\). Give your answer to the nearest degree.
M1 cosine rule for angle \(ABC\): \(\cos ABC = \frac{8^2 + BC^2 - 11^2}{2(8)BC}\) (or sine rule, noting \(ABC\) is acute because \(BAC\) is the largest angle)
✓ 1
A1 \(ABC = 64.1^\circ\)
✓ 1
M1 bearing of \(A\) from \(B\) is \(065^\circ + 180^\circ = 245^\circ\), so bearing of \(C\) is \(245^\circ - 64.1\ldots^\circ\)
✓ 1
A1 \(181^\circ\)
✓ 1
Common error: Bearings are measured clockwise from north: find the back bearing (add or subtract 180 degrees) before using an angle of the triangle.
258IBMA3S-3-18r40-073.7 Circular functions and their graphsPaper 2Medium7 marks
Calculator
In a coffee-roasting hopper, the depth of beans \(d\) cm, \(t\) minutes after the hopper starts to be refilled and emptied, is modelled by \(d(t) = 9 + 4\sin\left(\frac{\pi t}{6}\right)\) for \(0 \le t \le 12\), where the angle is in radians.
(a)
Write down the maximum depth and the minimum depth.
(2)
(b)
Find the period of the model.
(1)
(c)
Find the two times in the interval \(0 \le t \le 12\) when the depth is \(12\) cm. Give your answers to 3 significant figures.
M1 second solution uses \(\pi - 0.848\ldots\) (or symmetry about \(t = 3\))
✓ 1
A1 \(t = 4.38\)
✓ 1
Exam tip: Set your calculator to radians when the argument of a trigonometric function is written in terms of pi.
259IBMA3S-3-18r40-093.1 Three-dimensional geometryPaper 2Medium6 marks
Calculator
The roof of a coffee-roasting plant is a pyramid \(VABCD\) with a horizontal rectangular base $ABCD$, where \(AB = 6\) m and \(BC = 8\) m. The vertex \(V\) is \(7\) m vertically above the centre \(O\) of the base.
Diagram not to scale
(a)
Find the length \(AO\).
(2)
(b)
Find the length of the edge \(VA\). Give your answer to 3 significant figures.
(2)
(c)
Find the angle between the edge \(VA\) and the base $ABCD$. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(AC = \sqrt{6^2 + 8^2}\)
✓ 1
A1 \(AO = \frac{10}{2} = 5\) m
✓ 1
Part (b)
M1 \(VA = \sqrt{5^2 + 7^2}\)
✓ 1
A1 \(VA = 8.60\) m
✓ 1
Part (c)
M1 angle \(VAO\) with \(\tan(VAO) = \frac{7}{5}\)
✓ 1
A1 \(54.5^\circ\)
✓ 1
Common error: The angle between a line and a plane is measured in the vertical right-angled triangle, using the line and its projection onto the plane, not along a base edge.
260IBMA3S-3-18r40-133.4 Radian measure, arcs and sectorsPaper 2Medium12 marks
Calculator
In a roasting drum, the height \(h\) metres of the tip of a paddle above the floor is modelled by \(h(t) = 0.9 - 0.5\cos(72t)^\circ\), where \(t\) is the time in seconds and the angle is measured in degrees.
(a)
Write down the minimum and the maximum height of the paddle tip.
(2)
(b)
Find the time taken for one complete turn of the drum.
(1)
(c)
Find the first time at which the paddle tip is \(1.2\) m above the floor. Give your answer to 3 significant figures.
(3)
(d)
Find the second time, in the same turn, at which the paddle tip is \(1.2\) m above the floor. Give your answer to 3 significant figures.
(2)
(e)
Find the length of time in each turn for which the paddle tip is above \(1.2\) m. Use unrounded values and give your answer to 3 significant figures.
(2)
(f)
Find the percentage of each turn for which the paddle tip is above \(1.2\) m. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 minimum \(0.4\) m
✓ 1
A1 maximum \(1.4\) m
✓ 1
Part (b)
A1 \(\frac{360}{72} = 5\) s
✓ 1
Part (c)
M1 \(0.9 - 0.5\cos(72t) = 1.2\), so \(\cos(72t) = -0.6\)
Common error: The second solution of \(\cos\theta = k\) in one turn is \(360^\circ - \theta\), not \(180^\circ - \theta\) (which belongs to \(\sin\theta = k\)).
261IBMA3S-3-17r48-023.3 Applications of trigonometryPaper 2Medium7 marks
Calculator
From a mountain hut H, a climber walks 3.4 km to a summit A. A second climber walks 5.1 km from H to a summit B. The angle AHB is 72°.
Diagram not to scale
(a)
Find the distance AB, giving your answer in km to 3 significant figures.
(3)
(b)
Find the area of triangle AHB, giving your answer in km² to 3 significant figures.
(2)
(c)
The path from A to B is straight. Find the shortest distance from H to this path, in km to 3 significant figures. Use unrounded values from earlier parts.
M1 attempt to use area \(= \frac{1}{2}\times AB\times d\), so \(d = \frac{2\times\text{area}}{AB}\)
✓ 1
A1 distance = 3.18 km (ECF from (a) and (b))
✓ 1
Exam tip: When two sides and the included angle are known, the cosine rule gives the third side and the sine-based area formula gives the area directly.
262IBMA3S-3-17r48-033.5 The unit circle and the ambiguous casePaper 2Medium7 marks
Calculator
A telecoms company is siting a relay station R. In triangle ART, where A and T are existing towers, angle TAR = 50°, AT = 10 km and RT = 8.5 km. There are two possible positions for R that satisfy these conditions.
Diagram not to scale
(a)
Find the two possible sizes of angle ART, giving each to 1 decimal place.
(4)
(b)
Hence find the smaller of the two possible distances AR, in km to 3 significant figures. Use unrounded angles.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use the sine rule: \(\frac{\sin ART}{10} = \frac{\sin 50°}{8.5}\)
✓ 1
A1 acute angle ART = 64.3°
✓ 1
M1 use of $180° - $ acute angle for the second solution
✓ 1
A1 obtuse angle ART = 115.7°
✓ 1
Part (b)
M1 angle ATR for the smaller case \(= 180° - 50° - 115.7° = 14.3°\)
✓ 1
M1 attempt to use the sine rule: \(AR = \frac{8.5\sin(ATR)}{\sin 50°}\)
✓ 1
A1 AR = 2.74 km
✓ 1
Common error: After finding one angle with the sine rule, check whether 180° minus that angle also fits in the triangle; the sine rule alone cannot tell the two apart.
263IBMA3S-3-17r48-063.4 Radian measure, arcs and sectorsPaper 2Medium8 marks
Calculator
A circular pie tin has radius 9 cm. A straight cut across the pie makes a chord that subtends an angle of 2.2 radians at the centre O of the tin. The smaller piece is a circular segment bounded by the chord and the arc.
Diagram not to scale
(a)
Find the length of the arc of the smaller piece.
(2)
(b)
Find the area of the smaller piece, in cm² to 3 significant figures.
(3)
(c)
Find the perimeter of the smaller piece, in cm to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use \(l = r\theta\)
✓ 1
A1 \(l = 9\times 2.2 = 19.8\) cm
✓ 1
Part (b)
M1 area of sector \(= \frac{1}{2}(9)^2(2.2) = 89.1\)
✓ 1
M1 area of triangle \(= \frac{1}{2}(9)^2\sin 2.2 = 32.74\) and subtract
✓ 1
A1 area = 56.4 cm²
✓ 1
Part (c)
M1 attempt to find the chord, e.g. \(2(9)\sin\left(\frac{2.2}{2}\right)\) or the cosine rule
✓ 1
A1 chord = 16.04 cm
✓ 1
A1 perimeter = 19.8 + 16.04 = 35.8 cm
✓ 1
Exam tip: Set your calculator to radians for the whole question, and use sector minus triangle for the area of a segment.
264IBMA3S-3-17r48-083.7 Circular functions and their graphsPaper 2Medium8 marks
Calculator
The temperature \(T\) °C in a cheese-ripening room is modelled by \(T = 12 + 3\sin\left(\frac{\pi t}{12}\right)\), where \(t\) is the time in hours after midnight, \(0 \le t \le 24\). The argument of the sine is in radians.
(a)
Write down the maximum temperature and the period of the model.
(2)
(b)
The temperature first reaches 13.8 °C at time \(t_1\) and falls back to 13.8 °C at time \(t_2\). Find \(t_1\) and \(t_2\), giving each to 3 significant figures.
(4)
(c)
Find the number of hours in this day for which the temperature is above 13.8 °C, to 3 significant figures. Use unrounded values of \(t_1\) and \(t_2\).
M1 attempt to use area \(= \frac{1}{2}\times AB\times h\)
✓ 1
A1 \(h = 5.62\) km (ECF from (b))
✓ 1
Exam tip: Use the cosine rule for an angle when all three sides are known, then reuse that angle in the sine-area formula.
266AASL3-q00083.1 Three-dimensional geometryPaper 2Hard9 marks
Diagram not to scale
The diagram shows a cuboid \(OABCDEFG\) in which \(OA=8\) cm, \(OC=6\) cm and \(OD=4\) cm. Coordinates are defined so that \(O\) is the origin and \(A\), \(C\) and \(D\) have coordinates \((8,0,0)\), \((0,6,0)\) and \((0,0,4)\) respectively. The point \(M\) is the midpoint of the edge \([EF]\).
(a)
Write down the coordinates of \(M\).
(1)
(b)
Find the length \(OM\).
(2)
(c)
Find the angle between the line \([OM]\) and the base \(OABC\).
(3)
(d)
Find the size of the angle \(M\hat{O}E\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(M(8,\,3,\,4)\)
A1
(b)
\(\sqrt{8^{2}+3^{2}+4^{2}}\)
(M1)
FT from their \(M\).
\(\sqrt{89}=9.43\) cm
A1
\(9.43398\ldots\)
(c)
identifying the right-angled triangle \(ONM\), where \(N(8,3,0)\) is vertically below \(M\)
(M1)
May be seen on the diagram.
\(\sin\theta=\dfrac{4}{\sqrt{89}}\) or \(\tan\theta=\dfrac{4}{\sqrt{73}}\)
(M1)
\(25.1^{\circ}\)
A1
\(25.0873\ldots^{\circ}\); \(0.438\) radians
(d)
\([ME]\) is perpendicular to the face \(OAED\), so triangle \(OEM\) has a right angle at \(E\)
(R1)
Or finding all three sides \(OE=\sqrt{80}\), \(EM=3\), \(OM=\sqrt{89}\) for the cosine rule.
\(\tan M\hat{O}E=\dfrac{3}{\sqrt{80}}\)
(M1)
Or \(\cos M\hat{O}E=\dfrac{80+89-9}{2\sqrt{80}\sqrt{89}}\).
267AASL3-q00123.1 Three-dimensional geometryPaper 2Hard9 marks
Diagram not to scale
The diagram shows a wedge \(ABCDEF\). The base \(ABCD\) is a horizontal rectangle with \(AB=12\) cm and \(BC=5\) cm. The face \(DCFE\) is a vertical rectangle with \(DE=CF=3\) cm. The sloping face \(ABFE\) is a rectangle, and its diagonals \([AF]\) and \([BE]\) are drawn.
(a)
Find the length \(AF\).
(2)
(b)
Find the angle between the line \([AF]\) and the base \(ABCD\).
(2)
(c)
Find the angle between the line \([AE]\) and the base \(ABCD\).
(2)
(d)
Find the acute angle between the diagonals \([AF]\) and \([BE]\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(AF=\sqrt{12^{2}+5^{2}+3^{2}}\)
(M1)
Or \(AC=13\) then \(\sqrt{13^{2}+3^{2}}\).
\(\sqrt{178}=13.3\) cm
A1
\(13.3416\ldots\)
(b)
\(F\) is vertically above \(C\): \(\sin F\hat{A}C=\dfrac{3}{\sqrt{178}}\) or \(\tan F\hat{A}C=\dfrac{3}{13}\)
(M1)
\(13.0^{\circ}\)
A1
\(12.9946\ldots^{\circ}\); \(0.227\) radians
(c)
\(E\) is vertically above \(D\): \(\tan E\hat{A}D=\dfrac{3}{5}\)
(M1)
\(31.0^{\circ}\)
A1
\(30.9637\ldots^{\circ}\); \(0.540\) radians
(d)
\(AE=\sqrt{34}\;(=5.83)\); the diagonals of rectangle \(ABFE\) bisect each other at \(X\)
(M1)
triangle \(AXE\) is isosceles; \(\tan\left(\tfrac{1}{2}A\hat{X}E\right)=\dfrac{\sqrt{34}/2}{6}\), so \(A\hat{X}E=2\arctan\dfrac{\sqrt{34}}{12}\)
(M1)
Or the cosine rule in triangle \(AXE\) with \(AX=XE=\dfrac{\sqrt{178}}{2}\).
\(51.8^{\circ}\)
A1
\(51.8314\ldots^{\circ}\); \(0.905\) radians. Award A0 for \(128^{\circ}\) only.
268AASL3-q00143.1 Three-dimensional geometryPaper 2Hard16 marks
Diagram not to scale
A bucket has the shape of a frustum of a right circular cone: a cone with its tip removed by a cut parallel to its base. The bottom of the bucket is a horizontal circle of radius \(5\) cm, the open top is a horizontal circle of radius \(10\) cm, and the depth of the bucket is \(12\) cm. When the sloping sides are extended they meet at the point \(V\), vertically below the centre of the bottom.
Ignore the thickness of the material.
(a)
Show that \(V\) is \(12\) cm below the bottom of the bucket.
(2)
(b)
Show that the volume of the bucket is \(700\pi\text{ cm}^{3}\).
(3)
(c)
Write down the capacity of the bucket in litres.
(1)
(d)
The outside of the bucket, including the bottom, is painted. Find the area that is painted.
(4)
(e)
The bucket contains \(1.5\) litres of water. Find the depth of the water.
(4)
(f)
Find the angle between a sloping side of the bucket and the horizontal.
(2)
Show mark scheme
Working
Mark
Notes
(a)
similar triangles: \(\dfrac{h}{5}=\dfrac{h+12}{10}\), where \(h\) is the distance of \(V\) below the bottom
269AASL3-q00223.2 Right-angled and non-right-angled trigonometryPaper 2Hard11 marks
In triangle \(ABC\), the point \(D\) lies on \([AC]\). It is given that \(AB=9\) cm, \(B\hat{A}D=65^{\circ}\), \(A\hat{B}D=52^{\circ}\) and \(DC=5\) cm.
Or the cosine rule in triangle \(ABC\) or \(BDC\). FT from (a), (b).
\(41.7^{\circ}\)
A1
\(41.6965\ldots^{\circ}\). The angle is acute because triangle \(BDC\) already contains the obtuse angle \(117^{\circ}\).
Answers: (a) \(9.15\) cm · (b) \(12.3\) cm · (c) \(52.9\text{ cm}^{2}\) · (d) \(41.7^{\circ}\)
270AASL3-q00283.3 Applications of trigonometryPaper 2Hard10 marks
The top of a lighthouse, \(L\), is \(60\) m vertically above sea level, and the base of the lighthouse is at \(O\). Two boats, \(A\) and \(B\), are on the sea. From \(L\), boat \(A\) is on a bearing of \(220^{\circ}\) with an angle of depression of \(8^{\circ}\), and boat \(B\) is on a bearing of \(295^{\circ}\) with an angle of depression of \(14^{\circ}\).
Plan view, not to scale
(a)
Find the horizontal distance \(OA\).
(2)
(b)
Find the horizontal distance \(OB\).
(1)
(c)
Find the distance between the two boats.
(3)
(d)
Find the bearing of \(B\) from \(A\).
(4)
Show mark scheme
Working
Mark
Notes
(a)
\(\tan8^{\circ}=\dfrac{60}{OA}\)
(M1)
Angle of elevation of L from A equals the angle of depression (alternate angles).
bearing of \(O\) from \(A\) is \(220^{\circ}-180^{\circ}=040^{\circ}\), and \(B\) lies anticlockwise of \([AO]\) as seen from \(A\)
(M1)
A correct diagram showing the north line at A is sufficient.
\(007.5^{\circ}\)
A1
\(7.48358\ldots^{\circ}\) Accept \(007^{\circ}\) or \(7.48^{\circ}\).
Answers: (a) \(427\) m · (b) \(241\) m · (c) \(432\) m · (d) \(007.5^{\circ}\)
271AASL3-q00313.3 Applications of trigonometryPaper 2Hard9 marks
A straight hillside slopes upwards at \(18^{\circ}\) to the horizontal. A vertical tree \([BT]\) grows on the hillside with its base at \(B\), where \(B\) is \(25\) m from the point \(P\) at the foot of the slope, measured along the slope. The point \(P\), the base \(B\) and the top of the tree \(T\) lie in the same vertical plane. The angle of elevation of \(T\) from \(P\) is \(48^{\circ}\).
(a)
Draw a clearly labelled diagram to represent this information.
(1)
(b)
Show that \(P\hat{B}T=108^{\circ}\).
(2)
(c)
Find the height of the tree, \(BT\).
(3)
(d)
Find the height of \(T\) above the horizontal through \(P\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
horizontal through \(P\); \([PB]\) at \(18^{\circ}\) labelled \(25\) m; vertical \([BT]\); \([PT]\) at \(48^{\circ}\) to the horizontal
A1
(b)
\([BP]\) makes an angle of \(18^{\circ}\) below the horizontal through \(B\), and \([BT]\) makes \(90^{\circ}\) above it
M1
Or: the slope makes \(90^{\circ}-18^{\circ}=72^{\circ}\) with the vertical on the uphill side, and \(P\hat{B}T=180^{\circ}-72^{\circ}\).
\(P\hat{B}T=90^{\circ}+18^{\circ}\)
A1
\(P\hat{B}T=108^{\circ}\)
AG
(c)
\(B\hat{P}T=48^{\circ}-18^{\circ}=30^{\circ}\) and \(P\hat{T}B=180^{\circ}-30^{\circ}-108^{\circ}=42^{\circ}\)
\(35.5332\ldots\) Or height of \(B\) above \(P\) \(=25\sin18^{\circ}=7.73\).
\(35.53\ldots\times\sin48^{\circ}\)
(M1)
Or \(7.725\ldots+18.68\ldots\).
\(26.4\) m
A1
\(26.4063\ldots\)
Answers: (c) \(18.7\) m · (d) \(26.4\) m
Model sketch
Model diagram, not to scale
272AASL3-q00323.3 Applications of trigonometryPaper 2Hard14 marks
A yacht leaves port \(P\) and sails \(18\) km on a bearing of \(050^{\circ}\) to a buoy \(A\). It then sails \(25\) km on a bearing of \(130^{\circ}\) to a lighthouse \(B\).
(a)
Draw a clearly labelled diagram to represent this information.
(2)
(b)
Show that \(P\hat{A}B=100^{\circ}\).
(2)
(c)
Find the distance \(PB\).
(3)
(d)
Find the bearing of \(B\) from \(P\).
(3)
(e)
Write down the bearing of \(P\) from \(B\).
(1)
(f)
Find the shortest distance from \(A\) to the line \((PB)\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
north lines at \(P\) and \(A\); \([PA]\) labelled \(18\) km at \(50^{\circ}\) clockwise from north at \(P\)
A1
\([AB]\) labelled \(25\) km at \(130^{\circ}\) clockwise from north at \(A\), and \([PB]\) drawn
A1
(b)
bearing of \(P\) from \(A\) is \(050^{\circ}+180^{\circ}=230^{\circ}\)
M1
Or co-interior angles: angle between north at \(A\) and \([AP]\) is \(130^{\circ}\).
Accept \(098^{\circ}\). FT \(50^{\circ}+\) their angle.
(e)
\(277.8^{\circ}\)
A1
Accept \(278^{\circ}\). FT their (d) \(+180^{\circ}\).
(f)
area of triangle \(PAB=\dfrac{1}{2}(18)(25)\sin100^{\circ}\;(=222)\)
(M1)
Or \(d=18\sin A\hat{P}B\).
\(\dfrac{1}{2}(33.245\ldots)d=221.58\ldots\)
(M1)
Or \(18\sin47.778\ldots^{\circ}\).
\(13.3\) km
A1
\(13.3299\ldots\)
Answers: (c) \(33.2\) km · (d) \(097.8^{\circ}\) · (e) \(277.8^{\circ}\) · (f) \(13.3\) km
Model sketch
Model diagram, not to scale
273AASL3-q00333.3 Applications of trigonometryPaper 2Hard16 marks
Diagram not to scale
The diagram shows a field \(ABCD\) in the shape of a quadrilateral. \(AB=85\) m, \(BC=60\) m, \(AD=95\) m, \(A\hat{B}C=108^{\circ}\) and \(C\hat{A}D=40^{\circ}\).
(a)
Find \(AC\).
(3)
(b)
Find the area of triangle \(ABC\).
(2)
(c)
Find \(CD\).
(3)
(d)
Find the area of the field.
(3)
(e)
A fence is to be built around the edge of the field. The fencing costs €18.50 per metre. Find the cost of the fence, giving your answer to the nearest euro.
(3)
(f)
A water pipe is to be laid in a straight line from \(C\) to the edge \([AD]\). Find the shortest possible length of the pipe.
Answers: (a) \(118\) m · (b) \(2430\text{ m}^{2}\) · (c) \(76.1\) m · (d) \(6030\text{ m}^{2}\) · (e) €5848 · (f) \(76.0\) m
274AASL3-q00343.3 Applications of trigonometryPaper 2Hard15 marks
Two fire-lookout towers, \(A\) and \(B\), stand on level ground, with \(B\) \(12\) km due east of \(A\). A fire is seen at point \(F\). From \(A\) the bearing of \(F\) is \(042^{\circ}\), and from \(B\) the bearing of \(F\) is \(316^{\circ}\).
(a)
Draw a clearly labelled diagram to represent this information.
(2)
(b)
Show that \(A\hat{F}B=86^{\circ}\).
(2)
(c)
Find the distance \(AF\).
(3)
(d)
Find the distance \(BF\).
(2)
(e)
A fire station \(C\) is \(5\) km due south of \(B\). Find the distance from \(C\) to the fire.
(3)
(f)
Find the bearing of \(F\) from \(C\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(A\) and \(B\) with \(B\) due east of \(A\), labelled \(12\) km, with north lines at \(A\) and \(B\)
A1
\(F\) north of \([AB]\), with the angles \(42^{\circ}\) at \(A\) and \(316^{\circ}\) (or \(44^{\circ}\) west of north) at \(B\) shown
A1
(b)
\(F\hat{A}B=90^{\circ}-42^{\circ}=48^{\circ}\) and \(F\hat{B}A=316^{\circ}-270^{\circ}=46^{\circ}\)
\(360^{\circ}-28.514\ldots^{\circ}=331.485\ldots^{\circ}\). Accept \(331^{\circ}\) or \(332^{\circ}\).
Answers: (c) \(8.65\) km · (d) \(8.94\) km · (e) \(13.0\) km · (f) \(331.5^{\circ}\)
Model sketch
Model diagram, not to scale
275AASL3-q00433.4 Radian measure, arcs and sectorsPaper 2Hard10 marks
Diagram not to scale
Two circles have centres \(A\) and \(B\) and radii \(6\) cm and \(8\) cm respectively, with \(AB=10\) cm. The circles intersect at \(P\) and \(Q\). The shaded region is the region common to both circles.
(a)
Show that \(A\hat{P}B\) is a right angle.
(1)
(b)
Find \(P\hat{A}Q\), in radians.
(2)
(c)
Find \(P\hat{B}Q\), in radians.
(1)
(d)
Find the area of the shaded region.
(4)
(e)
Find the perimeter of the shaded region.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(AP^{2}+BP^{2}=36+64=100=AB^{2}\), so by the converse of Pythagoras’ theorem \(A\hat{P}B=\dfrac{\pi}{2}\)
A1
\(A\hat{P}B\) is a right angle
AG
(b)
\(\cos P\hat{A}B=\dfrac{6}{10}\) and \(P\hat{A}Q=2\,P\hat{A}B\)
(M1)
By symmetry about (AB).
\(1.85\)
A1
\(1.85459\ldots\)
(c)
\(2\arccos\dfrac{8}{10}=1.29\)
A1
\(1.28700\ldots\) Or \(\pi-1.85459\ldots\).
(d)
segment of circle \(A\): \(\dfrac{1}{2}(6)^{2}\left(1.854\ldots-\sin1.854\ldots\right)\)
(M1)
\(16.1\)
(A1)
\(16.1026\ldots\)
segment of circle \(B\): \(\dfrac{1}{2}(8)^{2}\left(1.287\ldots-\sin1.287\ldots\right)\;(=10.46)\)
(M1)
Or sector \(A\) \(+\) sector \(B\) \(-\) kite \(APBQ\): \(33.38+41.18-48\).
277AASL3-q00553.5 The unit circle and the ambiguous casePaper 2Hard15 marks
A lighthouse \(L\) is \(9\) km from a harbour \(H\), on a bearing of \(035^{\circ}\) from \(H\). A ship leaves \(H\) and sails due north at a constant speed of \(15\) km h−1. The light from the lighthouse can be seen clearly from the ship only when the ship is within \(6.5\) km of \(L\).
Dashed circle: the 6.5 km visibility range. Diagram not to scale
(a)
Find the shortest distance between the ship’s path and \(L\).
(2)
(b)
Let \(S\) be a position of the ship at which \(SL=6.5\) km. Show that \(\sin H\hat{S}L=0.794\), correct to three significant figures.
(2)
(c)
Find the two possible values of \(H\hat{S}L\).
(3)
(d)
Hence find the two possible distances \(HS\).
(4)
(e)
Find the length of time, in minutes, for which the light can be seen clearly from the ship.
(2)
(f)
Find the bearing of \(L\) from the ship at the moment the light first becomes clearly visible.
the light first becomes visible at \(HS=3.42\) km, where \(H\hat{S}L=127.4^{\circ}\); bearing \(=180^{\circ}-127.4^{\circ}\)
(M1)
\(052.6^{\circ}\)
A1
Accept \(053^{\circ}\).
Answers: (a) 5.16 km · (c) \(52.6^{\circ}\), \(127.4^{\circ}\) · (d) 3.42 km, 11.3 km · (e) 31.6 min · (f) \(052.6^{\circ}\)
278AASL3-q00783.7 Circular functions and their graphsPaper 2Hard16 marks
A Ferris wheel has radius \(20\) m and its centre is \(23\) m above the ground. The wheel rotates at a constant speed and completes one rotation every \(12\) minutes. A passenger boards at the lowest point of the wheel at time \(t=0\). The height of the passenger above the ground, \(h\) metres, \(t\) minutes after boarding, is modelled by
\(h(t)=a\cos(bt)+d\), where \(b\gt0\).
(a)
Write down the minimum height and the maximum height of the passenger above the ground.
(2)
(b)
Find the value of \(a\), the value of \(b\) and the value of \(d\).
(5)
(c)
Find the height of the passenger \(2.5\) minutes after boarding.
(2)
(d)
During one complete rotation, find the length of time for which the passenger is more than \(35\) m above the ground.
(4)
(e)
A second passenger boards at the lowest point exactly \(3\) minutes after the first passenger. Find the first time \(t\), where \(t\gt3\), at which the two passengers are at the same height above the ground, and find this height.
\(-20\cos\left(\dfrac{\pi t}{6}\right)+23=35\), or graph of \(h\) with \(y=35\)
(M1)
\(t=4.23\)
(A1)
\(4.22899\ldots\)
\(t=7.77\)
(A1)
\(7.77100\ldots\)
\(3.54\) minutes
A1
\(3.54200\ldots\)
(e)
\(h(t)=h(t-3)\)
(M1)
Second passenger’s height is \(h(t-3)\); GDC intersection.
\(t=7.5\) minutes
A1
\(37.1\) m
A1
\(37.1421\ldots\;(=23+10\sqrt{2})\)
Answers: (a) \(3\) m, \(43\) m · (b) \(a=-20,\ b=\dfrac{\pi}{6},\ d=23\) · (c) \(17.8\) m · (d) \(3.54\) minutes · (e) \(t=7.5\), \(37.1\) m
279AASL3-q00793.7 Circular functions and their graphsPaper 2Hard15 marks
Two buoys, \(A\) and \(B\), move up and down on the waves. Their heights, in metres, above the sea bed \(t\) seconds after a recording starts are modelled by
\(h_{A}(t)=1.4\sin(1.3t)+2.0\) and \(h_{B}(t)=0.9\cos(0.8t)+2.1\), for \(0\le t\le10\).
The graphs of \(h_{A}\) (solid) and \(h_{B}\) (dashed) are shown.
(a)
Write down the maximum height of buoy \(A\).
(1)
(b)
Find the period of \(h_{A}\).
(2)
(c)
Find the first time at which buoy \(A\) is at its maximum height.
(2)
(d)
Find the values of \(t\) for which the two buoys are at the same height.
(3)
(e)
Find the total time, during the 10 seconds, for which buoy \(A\) is higher than buoy \(B\).
(3)
(f)
Find the greatest vertical distance between the two buoys during the 10 seconds.
(2)
(g)
The recording continues beyond \(t=10\) and the model remains valid. Find the first time after \(t=10\) at which buoy \(B\) is at its maximum height.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(3.4\) m
A1
(b)
\(\dfrac{2\pi}{1.3}\)
(M1)
\(4.83\) s
A1
\(4.83321\ldots\)
(c)
\(1.3t=\dfrac{\pi}{2}\)
(M1)
Or use of GDC maximum.
\(t=1.21\) s
A1
\(1.20830\ldots\)
(d)
\(h_{A}(t)=h_{B}(t)\), e.g. intersections on GDC
(M1)
at least three correct values from \(0.548,\ 2.61,\ 4.41,\ 6.84,\ 9.75\)
281IBMA2S-048-05-123.8 Trigonometric equationsPaper 2Hard14 marks
Calculator
The height, \(h\) metres, of a seat on a fairground ride above the ground is modelled by \(h(t)=20+6\cos 2t+8\cos t\), where \(t\) is the time in seconds, \(0\le t\le 2\pi\).
(a)
Show that \(h(t)=12\cos^2t+8\cos t+14\).
(2)
(b)
Find the height of the seat when \(t=0\).
(1)
(c)
Find the exact value of the minimum height of the seat. Find also the values of \(t\) at which this minimum occurs, giving your answers correct to 3 significant figures.
(5)
(d)
Find the values of \(t\) for which the seat is 20 m above the ground.
M1 treats \(h\) as a quadratic in \(c=\cos t\) and finds its vertex, e.g. \(c=-\frac{8}{2(12)}\), or solves \(\frac{dh}{dt}=-8\sin t(3\cos t+1)=0\) and tests the stationary points
✓ 1
A1 \(\cos t=-\frac13\) (lies within \(-1\le\cos t\le1\))
✓ 1
A1 minimum height \(12\left(\frac19\right)-\frac83+14=\frac{38}{3}\) m
✓ 1
A1 \(t=1.91\) s (accept \(\arccos\left(-\frac13\right)\))
✓ 1
A1 \(t=4.37\) s (accept \(2\pi-\arccos\left(-\frac13\right)\))
✓ 1
Part (d)
M1 sets \(12\cos^2t+8\cos t+14=20\)
✓ 1
A1 \(6\cos^2t+4\cos t-3=0\)
✓ 1
A1 \(\cos t=\frac{-4+\sqrt{88}}{12}=0.448\ldots\)
✓ 1
R1 rejects \(\cos t=-1.12\) since \(|\cos t|\le1\)
✓ 1
A1 \(t=1.11\) s
✓ 1
A1 \(t=5.18\) s
✓ 1
Common error: Solving \(\frac{dh}{dt}=0\) only via \(\sin t=0\) gives \(t=\pi\) and misses the true minimum. Treating \(h\) as a quadratic in \(\cos t\) shows that the minimum occurs at \(\cos t=-\frac13\), not at \(\cos t=-1\).
282IBMA2S-048-07-113.2 Right-angled and non-right-angled trigonometryPaper 2Hard12 marks
Calculator
On an athletics track, each bend is a semicircle and all lanes are 1.22 m wide. Runners' distances are measured along a path 0.30 m outside the inner edge of each lane. The inner edge of lane 1 has radius 36.5 m. The straights have the same length in every lane, so runners in outer lanes start further ahead (a 'stagger') to make every race the same length.
(a)
Find the distance run on one bend in lane 1.
(2)
(b)
Find the distance run on one bend in lane 5.
(2)
(c)
Show that, on one bend, a runner in any lane runs 1.22π m further than a runner in the lane immediately inside, regardless of which lanes they are in.
(3)
(d)
A 400 m race includes two bends. Find the stagger between lane 1 and lane 8 for this race.
(2)
(e)
The lane 8 runner's stagger is measured along the first bend of lane 8. Find the angle, in radians, that this stagger subtends at the centre of the bend.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 radius 36.5 + 0.30 = 36.8 m, distance = 36.8π
✓ 1
A1 116 m (115.6…)
✓ 1
Part (b)
M1 radius 36.8 + 4 × 1.22 = 41.68 m
✓ 1
A1 41.68π = 131 m (130.9…)
✓ 1
Part (c)
M1 lane radii R and R + 1.22 give bend lengths πR and π(R + 1.22)
✓ 1
A1 difference π(R + 1.22) − πR = 1.22π (≈ 3.83 m)
✓ 1
R1 the expression does not contain R, so the difference is the same for every pair of adjacent lanes
✓ 1
Part (d)
M1 7 lane gaps × 2 bends × 1.22π (= 17.08π)
✓ 1
A1 53.7 m
✓ 1
Part (e)
M1 lane 8 radius 36.8 + 7 × 1.22 = 45.34 m
✓ 1
M1 θ = 53.66 ÷ 45.34
✓ 1
A1 1.18 rad
✓ 1
Exam tip: Write the general lane radius algebraically before you substitute numbers. The 'show that' then reduces to one line of simplification.
283IBMA2S-048-07-123.4 Radian measure, arcs and sectorsPaper 2Hard10 marks
Calculator
A tight drive belt passes around two pulleys in a machine, without crossing itself. The pulleys have radii 8 cm and 3 cm, and their centres P and Q are 13 cm apart. Each straight section of the belt is a common tangent to both pulleys, touching the larger pulley at A and the smaller pulley at B. Let α be the angle between PA and PQ.
Diagram not to scale
(a)
Show that the length of each straight section of the belt is 12 cm.
(2)
(b)
Find the value of α in radians.
(2)
(c)
Find the angle, in radians, through which the belt is in contact with the larger pulley.
(2)
(d)
Find the total length of the belt.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 PA and QB are both perpendicular to AB, so they are parallel. Forms a right-angled triangle with hypotenuse PQ = 13 and one side 8 − 3 = 5
✓ 1
A1 AB = √(13² − 5²) = √144 = 12
✓ 1
Part (b)
M1 cos α = 5/13
✓ 1
A1 α = 1.18 rad (1.176…)
✓ 1
Part (c)
M1 2π − 2α
✓ 1
A1 3.93 rad (3.931…)
✓ 1
Part (d)
M1 arc on larger pulley 8 × 3.931… (= 31.45)
✓ 1
M1 arc on smaller pulley 3 × 2α (= 7.056)
✓ 1
M1 adds the two straight sections, 2 × 12
✓ 1
A1 62.5 cm
✓ 1
Common error: The belt does not wrap half-way round each pulley. When the radii differ, it covers more than a semicircle on the larger pulley and less than a semicircle on the smaller one.
284IBMA2S-048-02-093.5 The unit circle and the ambiguous casePaper 2Hard13 marks
Calculator
A Mars rover leaves its lander L and drives in a straight line. A radio beacon B is 5.0 km from L, and the rover's path makes an angle of 35° with LB. The rover can communicate with the beacon only when it is within 3.5 km of B. Let d km be the distance the rover has travelled from L when it is exactly 3.5 km from B.
Dashed circle: range of the beacon. Diagram not to scale
(a)
Show that d satisfies d2 − (10 cos 35°)d + 12.75 = 0.
(2)
(b)
Find the distances from L at which the rover enters and leaves the beacon's range.
(3)
(c)
Find the length of the path along which the rover is in range.
(2)
(d)
Let E be the point where the rover enters the range. Find angle LEB, justifying your choice.
(4)
(e)
Find the angle subtended at B by the part of the path along which the rover is in range.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 cosine rule: 3.52 = 52 + d2 − 2(5)d cos 35°
✓ 1
A1 12.25 = 25 + d2 − 10d cos 35°, rearranged to the given equation
✓ 1
Part (b)
M1 attempt to solve the quadratic, e.g. d = [10 cos 35° ± √((10 cos 35°)2 − 51)]/2
✓ 1
A1 enters at d = 2.09 km
✓ 1
A1 leaves at d = 6.10 km
✓ 1
Part (c)
M1 6.102 − 2.089
✓ 1
A1 4.01 km
✓ 1
Part (d)
M1 sine rule: sin(LEB) = 5 sin 35° / 3.5
✓ 1
A1 sin(LEB) = 0.8194
✓ 1
A1 LEB = 55.0° or 125.0°
✓ 1
R1 LE = 2.09 < EB = 3.5, so angle LBE < 35°, hence LEB > 110°; LEB = 125.0°
✓ 1
Part (e)
M1 triangle formed by B and the entry and exit points is isosceles with base angles 180° − 125.0° = 55.0°
✓ 1
A1 180° − 2(55.02°) = 70.0°
✓ 1
Exam tip: The two roots of the cosine-rule quadratic are the two ambiguous-case positions; the same ambiguity appears in the sine rule, and side lengths tell you which angle belongs to which point.
285IBMA2S-048-03-123.6 Trigonometric identitiesPaper 2Hard13 marks
Calculator
A rugby ball is kicked from the ground with speed 25 m s−1 at an angle θ above the horizontal, where 0° < θ < 90°. Air resistance is ignored. The horizontal distance travelled before the ball lands is modelled by \(R = \frac{25^2\sin 2\theta}{9.8}\) metres. When the ball has travelled a horizontal distance x metres, its height is modelled by \(y = x\tan\theta - \frac{9.8x^2}{2(25^2)\cos^2\theta}\) metres.
(a)
Find the maximum value of R and the value of θ at which it occurs.
(2)
(b)
Find the two values of θ for which the ball lands 50 m away.
(4)
(c)
The ball must pass a point 30 m horizontally from the kick at a height of exactly 2.44 m. Use \(\sin^2\theta + \cos^2\theta = 1\) to show that \(T = \tan\theta\) satisfies \(7.056T^2 - 30T + 9.496 = 0\).
M1 attempts to solve the quadratic, e.g. \(T = \frac{30 \pm \sqrt{900 - 4(7.056)(9.496)}}{2(7.056)}\)
✓ 1
A1 \(T = 0.3444\) or \(T = 3.907\)
✓ 1
M1 uses \(\theta = \tan^{-1} T\)
✓ 1
A1 θ = 19.0° and θ = 75.6°
✓ 1
Exam tip: Writing \(\frac{1}{\cos^2\theta}\) in terms of \(\tan\theta\) turns the equation into a quadratic in a single unknown. Both roots of this quadratic are valid, because each gives an angle between 0° and 90°.
286IBMA2S-048-00-113.1 Three-dimensional geometryPaper 2Hard13 marks
Calculator
The glass roof of an airport atrium is a right pyramid VABCD with a horizontal rectangular base ABCD, where AB = 8 m and BC = 6 m. The apex V is 12 m vertically above M, the centre of the base.
Diagram not to scale
(a)
Find the distance AM.
(2)
(b)
Find the length of the sloping edge VA.
(2)
(c)
Find the angle between VA and the base ABCD.
(2)
(d)
Find the angle AVB.
(3)
(e)
Find the total area of glass in the four triangular faces.
Common error: On a rectangular base the faces are not congruent. The face on the longer edge AB has its foot 3 m from M, which is half of BC, not half of AB.
287IBMA2S-048-01-103.4 Radian measure, arcs and sectorsPaper 2Hard15 marks
Calculator
The hub of a wind turbine is 80 m above the ground and each blade is 45 m long. The rotor turns at a constant 15 revolutions per minute. At \(t=0\) seconds, the tip of blade A is level with the hub and moving upwards. The height of the tip of blade A above the ground, \(h\) m, is modelled by \(h(t)=80+45\sin(bt)\).
(a)
Write down the maximum and minimum heights of the tip of blade A.
(2)
(b)
Show that \(b=\frac{\pi}{2}\).
(2)
(c)
A sensor records when the tip is more than 110 m above the ground. Find the length of time per revolution for which the tip of blade A is above 110 m.
(4)
(d)
The tip of blade B has height \(80+45\sin\left(\frac{\pi t}{2}+\frac{2\pi}{3}\right)\). Find the first time \(t>0\) at which the tips of blades A and B are at the same height, and state this height.
(4)
(e)
In stronger winds the rotor turns faster, at \(N\) revolutions per minute. The tip of blade A spends less than 0.8 seconds above 110 m in each revolution exactly when \(N\) is greater than a critical value. Find this critical value of \(N\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 maximum 125 m
✓ 1
A1 minimum 35 m
✓ 1
Part (b)
M1 period \(=\frac{60}{15}=4\) s
✓ 1
A1 \(b=\frac{2\pi}{4}=\frac{\pi}{2}\)
✓ 1
Part (c)
M1 \(80+45\sin\left(\frac{\pi t}{2}\right)=110\), so \(\sin\left(\frac{\pi t}{2}\right)=\frac{2}{3}\)
✓ 1
A1 \(t_1=0.465\) s
✓ 1
A1 \(t_2=2-t_1=1.54\) s
✓ 1
A1 \(t_2-t_1=1.07\) s
✓ 1
Part (d)
M1 equating \(\sin\left(\frac{\pi t}{2}\right)=\sin\left(\frac{\pi t}{2}+\frac{2\pi}{3}\right)\), or an equivalent graphical or GDC method
✓ 1
M1 using \(\sin\theta=\sin(\pi-\theta)\): \(\frac{\pi t}{2}+\frac{2\pi}{3}=\pi-\frac{\pi t}{2}\), or a correct intersection found on the GDC
✓ 1
A1 \(t=\frac{1}{3}\) s (0.333 s)
✓ 1
A1 102.5 m
✓ 1
Part (e)
M1 time above 110 m per revolution \(=\frac{\pi-2\arcsin(2/3)}{b}\), or a fraction 0.2677 of the period \(\frac{60}{N}\)
A1 \(N=\frac{60b}{2\pi}=20.1\) (20.08) revolutions per minute
✓ 1
Exam tip: Write the time above a level as a general expression in b first. This turns a 'how fast' question into a single equation or inequality.
288IBMA2S-048-06-113.3 Applications of trigonometryPaper 2Hard14 marks
Calculator
A drone T hovers vertically above a point F on horizontal ground. Two surveyors stand at points A and B on the ground, 200 m apart. Using a theodolite, they measure the horizontal angles FAB = 50° and FBA = 70°. The angle of elevation of T from A is 32°.
Diagram not to scale
(a)
Find the horizontal distance AF.
(3)
(b)
Find the height of the drone above the ground.
(2)
(c)
Find the angle of elevation of T from B.
(4)
(d)
Find the angle TAB between the line of sight from A to the drone and the line AB.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 angle AFB = 60°
✓ 1
M1 sine rule: \(\frac{AF}{\sin 70°} = \frac{200}{\sin 60°}\)
✓ 1
A1 AF = 217 m (217.0)
✓ 1
Part (b)
M1 \(h = AF\tan 32°\)
✓ 1
A1 \(h = 136\) m (135.6)
✓ 1
Part (c)
M1 sine rule for BF: \(\frac{BF}{\sin 50°} = \frac{200}{\sin 60°}\)
✓ 1
A1 BF = 177 m (176.9)
✓ 1
M1 \(\tan\theta = \frac{h}{BF}\)
✓ 1
A1 \(\theta = 37.5°\)
✓ 1
Part (d)
M1 attempt to find AT, e.g. \(AT = \frac{AF}{\cos 32°}\)
Alternative methods earn the same marks. For example, project T onto AB: \(AT\cos T\hat{A}B = AF\cos 50°\) with \(AF = AT\cos 32°\), giving \(\cos T\hat{A}B = \cos 32°\cos 50°\). A correct vector method using the scalar product also earns full marks.
—
Common error: Angle TAB lies in a sloping plane and is not equal to the horizontal angle FAB (50°) or to the elevation (32°). Identify the triangle that contains the required angle and find all its sides first.
289IBMA2S-048-04-113.3 Applications of trigonometryPaper 2Hard15 marks
Calculator
A rescue helicopter H hovers vertically above a point F on level ground. Two ground teams are at points P and Q, where Q is 2.4 km due east of P. From P, the bearing of F is 035°, and from Q, the bearing of F is 310°. From P, the angle of elevation of the helicopter is 14°.
Diagram not to scale
(a)
Show that angle PFQ = 85°.
(2)
(b)
Find the distance PF.
(3)
(c)
Find the height of the helicopter above the ground.
(2)
(d)
Find the angle of elevation of the helicopter from Q.
(4)
(e)
A third team is at M, the midpoint of PQ. Find the straight-line distance from M to the helicopter.
Exam tip: Work in the horizontal ground triangle first, using bearings to find the angles. Then treat the height as a vertical side of a separate right-angled triangle. Mixing the two planes in one sketch causes most errors.
290IBMA3S-3-17r6-123.1 Three-dimensional geometryPaper 2Hard11 marks
Calculator
A farmer models a grain store as a solid right pyramid VABCD. The base ABCD is a horizontal square of side 6 m with centre M, and the apex V is 4 m vertically above M. Let N be the midpoint of AB. Give angles to 1 decimal place and lengths to 3 significant figures.
Diagram not to scale
(a)
Find the length VN.
(2)
(b)
Find the angle between the face VAB and the base.
(2)
(c)
Find the volume of the store.
(2)
(d)
Find the length of the edge VA.
(2)
(e)
Find angle AVC.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 attempt to use Pythagoras in triangle VMN with MN = 3
✓ 1
A1 VN = 5 m
✓ 1
Part (b)
M1 \(\tan\theta=\frac43\) or \(\cos\theta=\frac35\)
✓ 1
A1 53.1°
✓ 1
Part (c)
M1 \(\frac13\times6^2\times4\)
✓ 1
A1 48 m3
✓ 1
Part (d)
M1 \(AM=\sqrt{3^2+3^2}\) (=\(3\sqrt2\)) and Pythagoras with VM
✓ 1
A1 \(VA=\sqrt{34}=5.83\) m
✓ 1
Part (e)
M1 attempt to use the cosine rule with \(AC=6\sqrt2\)
Exam tip: Draw the right-angled triangle containing the height of the pyramid for each length or angle, then switch to the cosine rule when the triangle is not right-angled.
291IBMA3S-3-17r6-133.3 Applications of trigonometryPaper 2Hard14 marks
Calculator
An aircraft leaves airport A and flies 120 km on a bearing of 065° to a waypoint B. It then flies 90 km on a bearing of 140° to a waypoint C. Use unrounded values in later parts; give distances to 1 decimal place and angles to 1 decimal place.
Diagram not to scale
(a)
Show that angle ABC = 105°.
(2)
(b)
Find the distance AC.
(3)
(c)
Find angle BAC, and hence the bearing of C from A.
(4)
(d)
Find the area of triangle ABC, to the nearest km2.
(2)
(e)
The aircraft flies the whole route A to B to C and back to A at a constant speed of 300 km/h. Find the total flight time in minutes, to the nearest minute.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 bearing of A from B is \(065^\circ+180^\circ=245^\circ\)
✓ 1
A1 \(245^\circ-140^\circ=105^\circ\) (AG)
✓ 1
Part (b)
M1 attempt to use the cosine rule
✓ 1
A1 \(AC^2=120^2+90^2-2(120)(90)\cos105^\circ\)
✓ 1
A1 AC = 167.6 km
✓ 1
Part (c)
M1 attempt to use the sine rule
✓ 1
A1 \(\sin BAC=\frac{90\sin105^\circ}{167.6}\)
✓ 1
A1 angle BAC = 31.2° (acute, as BC is the shorter side)
✓ 1
A1 bearing 065° + 31.2° = 096.2°
✓ 1
Part (d)
M1 \(\frac12(120)(90)\sin105^\circ\)
✓ 1
A1 5216 km2
✓ 1
Part (e)
M1 total distance \(120+90+167.6\)
✓ 1
M1 divide by 300 and convert to minutes
✓ 1
A1 76 minutes (75.5…)
✓ 1
Common error: A bearing is measured clockwise from north; find the interior angle by working with the back-bearing at the turning point.
292IBMA3S-3-17r8-123.5 The unit circle and the ambiguous casePaper 2Hard7 marks
Calculator
In a survey of an ash plume, triangle PQR has PQ = 8.4 km, PR = 6.1 km and angle PQR = 43°. Angle PRQ is not known, and two different triangles fit this data.
Diagram not to scale
(a)
Find the two possible sizes of angle PRQ, to the nearest tenth of a degree.
(3)
(b)
Hence find the two possible lengths of QR, to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sine rule \(\frac{\sin R}{8.4}=\frac{\sin43^\circ}{6.1}\)
✓ 1
A1 \(R=69.9^\circ\)
✓ 1
A1 \(R=180^\circ-69.9^\circ=110.1^\circ\)
✓ 1
Part (b)
M1 \(\angle QPR=180^\circ-43^\circ-R\) for each case, giving \(67.1^\circ\) and \(26.9^\circ\)
✓ 1
M1 sine rule \(QR=\frac{6.1\sin P}{\sin43^\circ}\)
✓ 1
A1 \(QR=8.24\) km
✓ 1
A1 \(QR=4.05\) km
✓ 1
Exam tip: If the angle you are finding comes from a sine, check whether \(180^\circ\) minus your answer also gives a valid triangle when the given side is shorter than the side opposite the known angle.
293IBMA3S-3-17r8-143.4 Radian measure, arcs and sectorsPaper 2Hard8 marks
Calculator
A circular revolving stage has centre O and radius 4.0 m. Chord AB subtends an angle \(\theta\) radians at O, where \(0<\theta<\pi\). The minor segment cut off by AB is to be a seating area.
Diagram not to scale
(a)
Show that the perimeter of the segment (the arc AB plus the chord AB) is \(4\theta+8\sin\frac{\theta}{2}\) m.
(3)
(b)
The perimeter of the seating area is 10.0 m. Find \(\theta\), to 3 significant figures.
(2)
(c)
Find the area of the seating area, to 3 significant figures. Use the unrounded value of \(\theta\) from (b).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 arc length \(=4\theta\)
✓ 1
M1 chord \(AB=2(4)\sin\frac{\theta}{2}\) (or by the cosine rule)
✓ 1
A1 sum gives \(4\theta+8\sin\frac{\theta}{2}\) (AG)
✓ 1
Part (b)
M1 equation \(4\theta+8\sin\frac{\theta}{2}=10\)
✓ 1
A1 \(\theta=1.29\) radians (from a graph or solver)
✓ 1
Part (c)
M1 area of sector minus triangle \(\frac12(4)^2(\theta-\sin\theta)\)
✓ 1
A1 substitute \(\theta=1.294\ldots\)
✓ 1
A1 \(2.66\) m\(^2\) (ECF from (b))
✓ 1
Exam tip: Use your calculator's equation solver or the intersection of two graphs; state the equation you are solving so the method mark is secure.
294IBMA3S-3-16r18-103.3 Applications of trigonometryPaper 2Hard13 marks
Calculator
Three wind turbines T1, T2 and T3 stand on level ground. The distance \(T_1T_2=420\) m, \(T_2T_3=510\) m and angle \(T_1T_2T_3=110^\circ\). Use unrounded values from earlier parts in later calculations. Give final answers to 3 significant figures.
Diagram not to scale
(a)
Find the distance \(T_1T_3\).
(3)
(b)
Find the size of angle \(T_2T_1T_3\).
(3)
(c)
Find the area of triangle \(T_1T_2T_3\).
(2)
(d)
Find the shortest distance from \(T_2\) to the straight line \(T_1T_3\).
(2)
(e)
Each tower is 95 m tall. A surveyor stands at the point on \(T_1T_3\) closest to \(T_2\). Find the angle of elevation from the surveyor to the top of the tower at \(T_2\).
M1 sine rule \(\frac{\sin T_2T_1T_3}{510}=\frac{\sin110^\circ}{T_1T_3}\)
✓ 1
A1 \(\sin T_2T_1T_3=0.6276\ldots\)
✓ 1
A1 \(38.9^\circ\)
✓ 1
Part (c)
M1 area \(=\frac{1}{2}(420)(510)\sin110^\circ\)
✓ 1
A1 \(1.01\times10^5\) m2 (100 641 m2 unrounded)
✓ 1
Part (d)
M1 area \(=\frac{1}{2}\times T_1T_3\times h\) with \(h\) the perpendicular distance
✓ 1
A1 \(h=264\) m
✓ 1
Part (e)
M1 \(\tan\theta=\frac{95}{h}\)
✓ 1
M1 substitute \(h=263.6\ldots\) (ECF from (d))
✓ 1
A1 \(\theta=19.8^\circ\)
✓ 1
Exam tip: When a triangle has one obtuse angle, find the other angles with the sine rule only after the obtuse angle is known, as the other two must be acute.
295IBMA3S-3-16r18-113.6 Trigonometric identitiesPaper 2Hard13 marks
Calculator
A wave pool produces waves so that the depth of water \(h\) cm at a fixed point is modelled by \(h(t)=a\cos(bt)+d\), where \(t\) is the time in seconds after the machine starts, \(b>0\) and the angle is in radians. The depth is greatest, 160 cm, when \(t=0\). The first time the depth is least, 80 cm, is when \(t=3\).
(a)
Find the values of \(a\), \(d\) and \(b\).
(4)
(b)
Find the values of \(t\), for \(0\le t\le6\), at which \(h=135\). Give your answers to 3 significant figures.
(4)
(c)
Hence find the time in each 6-second cycle for which the depth is greater than 135 cm. Give your answer to 3 significant figures.
(2)
(d)
Children can stand safely when the depth is at most 100 cm. Find, as an exact fraction, the proportion of each cycle for which this is true.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 \(a=40\)
✓ 1
A1 \(d=120\)
✓ 1
M1 period \(=2\times3=6\), so \(b=\frac{2\pi}{6}\)
✓ 1
A1 \(b=\frac{\pi}{3}\)
✓ 1
Part (b)
M1 \(120+40\cos\left(\frac{\pi t}{3}\right)=135\), so \(\cos\left(\frac{\pi t}{3}\right)=0.375\)
✓ 1
M1 \(\frac{\pi t}{3}=1.186\ldots\) (calculator in radian mode)
✓ 1
A1 \(t=1.13\)
✓ 1
A1 \(t=4.87\) (using \(6-1.133\ldots\))
✓ 1
Part (c)
M1 time \(=2\times1.133\ldots\) (or \(6-(4.867-1.133)\))
Common error: When solving trigonometric equations in radians, the second solution in a cycle comes from the symmetry of the graph, not from adding \(\pi\).
296IBMA3S-3-17r24-133.4 Radian measure, arcs and sectorsPaper 2Hard10 marks
Calculator
A sector-shaped stage OAB has radius r metres and angle θ radians at O. The perimeter of the stage is 24 m and its area is 32 m2.
Diagram not to scale
(a)
Show that r2 − 12r + 32 = 0.
(3)
(b)
Find the two possible values of r and the corresponding values of θ.
(3)
(c)
The designer needs the angle at O to be less than 2 radians. Write down the value of r that must be used.
(1)
(d)
For this stage, a straight curtain is hung along AB. Find the area of the segment between the curtain and the arc, to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 perimeter: 2r + rθ = 24, so rθ = 24 − 2r
✓ 1
M1 area: ½r(rθ) = 32, so ½r(24 − 2r) = 32
✓ 1
A1 12r − r2 = 32, so r2 − 12r + 32 = 0 (AG)
✓ 1
Part (b)
M1 solves the quadratic: (r − 4)(r − 8) = 0
✓ 1
A1 r = 4 or r = 8
✓ 1
A1 θ = 4 rad (r = 4) and θ = 1 rad (r = 8)
✓ 1
Part (c)
A1 r = 8
✓ 1
Part (d)
M1 area of triangle ½(8)2 sin 1 = 26.93 m2 (ECF from (c))
✓ 1
M1 segment = sector − triangle = 32 − 26.93...
✓ 1
A1 5.073 ≈ 5.07 m2
✓ 1
Common error: Both solutions of a quadratic may satisfy the algebra; always check each against the conditions in the question before choosing one.
297IBMA3S-3-17r24-143.3 Applications of trigonometryPaper 2Hard11 marks
Calculator
The summit T of a volcano is a height h km vertically above a point O on level ground. Observer A is due west of O and observer B is due south of O, so that angle AOB = 90°. The angle of elevation of T is 25° from A and 18° from B. The distance AB is 3.4 km. Give answers to 3 significant figures unless stated, and keep unrounded values in later parts.
Diagram not to scale
(a)
Write down expressions for OA and OB in terms of h.
(2)
(b)
Find the value of h.
(3)
(c)
Find the bearing of B from A, to the nearest degree.
(3)
(d)
Observer M stands at the midpoint of AB. Find the angle of elevation of T from M, to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 OA = h/tan 25°
✓ 1
A1 OB = h/tan 18°
✓ 1
Part (b)
M1 Pythagoras in triangle AOB: 3.42 = OA2 + OB2
✓ 1
M1 h2(1/tan2 25° + 1/tan2 18°) = 3.42 and solves for h
✓ 1
A1 h = 0.9064 ≈ 0.906 km
✓ 1
Part (c)
M1 tan(angle OAB) = OB/OA
✓ 1
A1 angle OAB = 55.1°
✓ 1
A1 bearing = 90° + 55.1° = 145.1° ≈ 145°
✓ 1
Part (d)
M1 OM = 1.7 km (the midpoint of the hypotenuse of a right-angled triangle is 1.7 km from the right angle, as the circle through A, O, B has diameter AB)
✓ 1
M1 tan(angle) = h/1.7
✓ 1
A1 28.07° ≈ 28.1°
✓ 1
Exam tip: Draw a plan view of the ground triangle and a separate vertical triangle for each observer; the height h links them.
298IBMA3S-3-17r25-103.1 Three-dimensional geometryPaper 2Hard14 marks
Calculator
The roof of a beehive is a pyramid VABCD. The base ABCD is a horizontal rectangle with AB = 80 cm and BC = 50 cm. The vertex V is vertically above the centre M of the base, and VM = 30 cm. Do not use rounded values in later parts; give final answers as stated.
Diagram not to scale
(a)
Show that VA = 55.9 cm, correct to 3 significant figures.
(4)
(b)
Find the angle between the edge VA and the base ABCD, giving your answer to 3 significant figures.
(3)
(c)
Find angle AVB, giving your answer to 3 significant figures.
(3)
(d)
The four triangular faces of the roof are to be covered with waterproof sheet. Calculate the total area of the four faces, to the nearest cm2.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use Pythagoras in triangle ABC: AC2 = 802 + 502 = 8900
✓ 1
M1 AM = \(\frac{1}{2}\)AC, so AM2 = 2225
✓ 1
M1 VA2 = 302 + 2225 = 3125
✓ 1
A1 VA = √3125 = 55.9 cm (AG)
✓ 1
Part (b)
M1 identify the required angle as angle VAM
✓ 1
M1 use tan VAM = \(\frac{30}{\sqrt{2225}}\)
✓ 1
A1 angle = 32.5°
✓ 1
Part (c)
M1 attempt to use the cosine rule in triangle AVB with VA = VB
✓ 1
A1 cos AVB = \(\frac{3125+3125-80^2}{2(3125)}\) = −0.024
✓ 1
A1 angle AVB = 91.4°
✓ 1
Part (d)
M1 attempt to find the area of triangle VAB, e.g. ½(VA)(VB) sin AVB or ½(80)(√1525)
✓ 1
A1 area VAB = 1562 cm2
✓ 1
M1 attempt to find the area of triangle VAD, e.g. ½(50)(√2500) = ½(50)(50)
✓ 1
A1 total area = 2(1562 + 1250) = 5624 cm2
✓ 1
Exam tip: In a pyramid, identify the right-angled triangle that contains the required angle (here the one with the vertical height) and draw it separately before using trigonometry.
299IBMA3S-3-17r25-143.7 Circular functions and their graphsPaper 2Hard10 marks
Calculator
The number of foraging bees in a colony, N thousand, is modelled by N(t) = a sin(b(t − c)) + d, for 0 ≤ t < 12, where t is the time in months after 1 January and a > 0, b > 0, 0 < c < 12 and the argument of sine is in radians. The number of foraging bees is greatest, 48 thousand, when t = 7 and least, 12 thousand, when t = 1. The pattern repeats every 12 months.
(a)
Find the values of a, b, c and d.
(6)
(b)
Find the total length of time, in months, during one year for which the number of foraging bees exceeds 42 thousand. Give your answer to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 a = \(\frac{48-12}{2}\) and d = \(\frac{48+12}{2}\)
✓ 1
A1 a = 18 and d = 30
✓ 1
M1 b = \(\frac{2\pi}{12}\)
✓ 1
A1 b = \(\frac{\pi}{6}\)
✓ 1
M1 sine is at its maximum when b(t − c) = \(\frac{\pi}{2}\), so 7 − c = 3
Exam tip: For an inequality on a sine model, find the two critical times in one cycle and subtract them: the sine curve is above the line only between them.
300IBMA3S-3-19r27-123.4 Radian measure, arcs and sectorsPaper 2Hard8 marks
Calculator
A circular pie dish has radius 15 cm. A chord AB of the circle subtends an angle \(\theta\) radians at the centre O, where \(0<\theta<\pi\). The area of the minor segment cut off by AB is 40 cm2.
Diagram not to scale
(a)
Show that \(\theta-\sin\theta=\frac{16}{45}\).
(3)
(b)
Solve the equation in part (a) to find \(\theta\), to 3 significant figures.
(2)
(c)
Find the perimeter of the segment, in cm to 3 significant figures. Use the unrounded value of \(\theta\).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 sector area \(\frac12\times15^2\times\theta\)
✓ 1
M1 triangle OAB area \(\frac12\times15^2\sin\theta\), subtracted from the sector
✓ 1
A1 \(112.5(\theta-\sin\theta)=40\), so \(\theta-\sin\theta=\frac{40}{112.5}=\frac{16}{45}\) (AG)
✓ 1
Part (b)
M1 solving numerically (graphing or equation solver)
Exam tip: An equation that mixes \(\theta\) and \(\sin\theta\) cannot be rearranged algebraically; solve it with the graphing or equation-solving function of your calculator.
301IBMA3S-3-19r27-133.3 Applications of trigonometryPaper 2Hard9 marks
Calculator
A climber is at camp A. Peak B is 5.2 km from A on a bearing of 040°. Peak C is 7.4 km from A on a bearing of 115°.
Diagram not to scale
(a)
Write down the size of angle BAC.
(1)
(b)
Find the distance BC, in km to 3 significant figures.
(2)
(c)
Find the size of angle ABC, to the nearest 0.1°. Use the unrounded value of BC.
(2)
(d)
Find the bearing of C from B, to the nearest degree. Use the unrounded value of angle ABC.
(2)
(e)
Find the shortest distance from A to the straight line BC, in km to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 115° − 40° = 75°
✓ 1
Part (b)
M1 \(BC^2=5.2^2+7.4^2-2(5.2)(7.4)\cos75°\)
✓ 1
A1 BC = 7.87 km
✓ 1
Part (c)
M1 \(\dfrac{\sin ABC}{7.4}=\dfrac{\sin75°}{BC}\)
✓ 1
A1 angle ABC = 65.3°
✓ 1
Part (d)
M1 the bearing of A from B is 220°, and the bearing of C from B is 220° − angle ABC
✓ 1
A1 155°
✓ 1
Part (e)
M1 \(5.2\sin(ABC)\)
✓ 1
A1 4.72 km
✓ 1
Exam tip: Draw a sketch with a north line at each point before using bearings; the back bearing is the forward bearing plus or minus 180°.
302IBMA3S-3-19r27-143.4 Radian measure, arcs and sectorsPaper 2Hard10 marks
Calculator
A communications satellite orbits directly above the equator at a height of 35 786 km above the Earth's surface. Model the Earth as a sphere of radius 6371 km with centre O. Let S be the satellite and let P be a point on the equator at the edge of the region from which S is visible, so that the line SP is a tangent to the Earth at P. Let angle POS = \(\theta\) radians.
Diagram not to scale
(a)
Show that \(\cos\theta=\dfrac{6371}{42157}\).
(2)
(b)
Find \(\theta\), in radians to 3 significant figures.
(2)
(c)
Find the length of the equator from which the satellite can be seen, in km to 3 significant figures. Use the unrounded value of \(\theta\).
(2)
(d)
A radio signal travels along SP at \(3.00\times10^{8}\) m s−1. Find the time taken for the signal to travel from S to P, in seconds to 3 significant figures.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R1 the tangent SP is perpendicular to the radius OP, so triangle OPS has a right angle at P
M1 time = distance in metres \(\div\,3.00\times10^{8}\)
✓ 1
A1 0.139 s
✓ 1
Common error: Convert kilometres to metres before dividing by a speed given in m s⁻¹; otherwise the time is wrong by a factor of 1000.
303IBMA3S-3-19r30-123.1 Three-dimensional geometryPaper 2Hard10 marks
Calculator
A drone D hovers at a height h metres vertically above a point F on level ground. Two observers stand at points A and B on the ground, with AB = 500 m and angle AFB = 70°. The angle of elevation of D from A is 35° and from B is 48°. Give final answers to 3 significant figures.
Diagram not to scale
(a)
Write expressions for AF and BF in terms of h.
(2)
(b)
Use the cosine rule in triangle AFB to show that h²[1⁄tan²35° + 1⁄tan²48° − 2cos 70° ⁄ (tan 35° tan 48°)] = 500².
(3)
(c)
Hence find h.
(3)
(d)
Find the distance AD, using the unrounded value of h.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 AF = h ⁄ tan 35°
✓ 1
A1 BF = h ⁄ tan 48°
✓ 1
Part (b)
M1 500² = AF² + BF² − 2(AF)(BF)cos 70°
✓ 1
A1 substitution of the expressions from (a)
✓ 1
A1 factorising h² to reach the given equation (AG)
✓ 1
Part (c)
M1 h = 500 ⁄ √[…]
✓ 1
A1 bracket = 1.97…
✓ 1
A1 h = 356 m
✓ 1
Part (d)
M1 AD = h ⁄ sin 35° (or Pythagoras in triangle AFD)
✓ 1
A1 621 m
✓ 1
Exam tip: When a 'show that' result is given, you may use it in the next part even if you could not prove it.
304IBMA3S-3-19r30-143.7 Circular functions and their graphsPaper 2Hard12 marks
Calculator
The temperature T °C in a fermentation tank is modelled by T = c + a cos(bt)°, where t is the time in hours since the tank was started and a > 0, b > 0. The temperature has its first maximum of 34 °C when t = 0 and its first minimum of 22 °C when t = 6.
(a)
Find the values of a, c and b.
(4)
(b)
The tank is at risk of over-fermenting when T ≥ 32. Find the first time at which T = 32, to 3 significant figures.
(3)
(c)
Find the total time in the first 12 hours for which T ≥ 32. Use unrounded values and give your answer to 3 significant figures.
(2)
(d)
A heater switches on whenever T < 25. Find the length of time in the first 12 hours for which the heater is on.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 a = 6
✓ 1
A1 c = 28
✓ 1
M1 half a period is 6 hours, so the period is 12
✓ 1
A1 b = 360 ÷ 12 = 30
✓ 1
Part (b)
M1 cos(30t) = 2⁄3
✓ 1
A1 30t = 48.19…°
✓ 1
A1 t = 1.61 hours
✓ 1
Part (c)
M1 the second time is 12 − 1.606… = 10.39… (by symmetry)
✓ 1
A1 3.21 hours
✓ 1
Part (d)
M1 cos(30t) = −½
✓ 1
A1 30t = 120° and 240°, so t = 4 and t = 8
✓ 1
A1 4 hours
✓ 1
Exam tip: Sketching the cosine curve from t = 0 to 12 shows at a glance which side of each solution satisfies an inequality.
305IBMA3S-3-19r31-123.2 Right-angled and non-right-angled trigonometryPaper 2Hard10 marks
Calculator
Two points A and B are 150 m apart on one straight bank of a river. A tree C stands on the opposite bank. Angle CAB = 48° and angle CBA = 71°. A point D lies on the line AB extended beyond B, with angle CDA = 25°. Treat the ground as flat. Give all lengths to the nearest metre and use unrounded values in later parts.
M1 sine rule in triangle BCD: BD/sin 46° = BC/sin 25°
✓ 1
A1 BD = 217 m
✓ 1
Exam tip: Label every angle in the diagram first; the angle at B in triangle BCD is the supplement of angle CBA, which is easy to overlook.
306IBMA3S-3-19r31-143.4 Radian measure, arcs and sectorsPaper 2Hard10 marks
Calculator
A coffee roaster's cooling tray is a sector of a circle of radius r cm and angle θ radians, where 0 < θ < 2π. The sector has perimeter 60 cm and area 200 cm².
Diagram not to scale
(a)
Show that θ = (60 − 2r)/r.
(2)
(b)
Hence show that r² − 30r + 200 = 0.
(3)
(c)
Find the two possible pairs of values of r and θ.
(3)
(d)
For the tray with the larger angle, find the straight-line distance between the two ends of the curved edge, correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 perimeter = 2r + rθ = 60
✓ 1
AG rearranges to θ = (60 − 2r)/r
✓ 1
Part (b)
M1 area = ½r²θ = 200
✓ 1
M1 substitutes: ½r(60 − 2r) = 200
✓ 1
AG 30r − r² = 200, so r² − 30r + 200 = 0
✓ 1
Part (c)
M1 factorises: (r − 10)(r − 20) = 0
✓ 1
A1 r = 10 and θ = 4
✓ 1
A1 r = 20 and θ = 1
✓ 1
Part (d)
M1 chord = 2 × 10 × sin(4/2)
✓ 1
A1 18.2 cm
✓ 1
Exam tip: Both roots of a quadratic must be checked against the context; here both satisfy 0 < θ < 2π, so both trays are possible.
307IBMA3S-3-18r40-103.8 Trigonometric equationsPaper 2Hard6 marks
Calculator
Solve the equation \(3\sin 2x = 2\cos x\) for \(0 \le x \le 2\pi\). Give any non-exact answers to 3 significant figures.
(a)
Find all the solutions.
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use \(\sin 2x = 2\sin x\cos x\)
✓ 1
M1 rearrange and factorise: \(2\cos x(3\sin x - 1) = 0\)
✓ 1
A1 \(\cos x = 0\) giving \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\)
✓ 1
A1 \(\sin x = \frac{1}{3}\)
✓ 1
A1 \(x = 0.340\)
✓ 1
A1 \(x = \pi - 0.339\ldots = 2.80\)
✓ 1
Common error: Dividing both sides by \(\cos x\) would lose the solutions where \(\cos x = 0\).
308IBMA3S-3-18r40-113.4 Radian measure, arcs and sectorsPaper 2Hard13 marks
Calculator
A footbridge has a circular arch. The straight horizontal span \(AB\) is \(40\) m long and the highest point of the arch is \(6\) m vertically above the midpoint \(M\) of \(AB\). The arch is part of a circle with centre \(O\) and radius \(r\) metres, and \(O\) is directly below \(M\).
Diagram not to scale
(a)
Show that \(r = 36.3\) m, correct to 3 significant figures.
(4)
(b)
Find the angle \(AOB\) in radians. Give your answer to 3 significant figures, using \(r = \frac{109}{3}\).
(2)
(c)
Find the length of the arch \(AB\) (the arc). Give your answer to 3 significant figures, using unrounded values.
(2)
(d)
Find the area of the region between the chord \(AB\) and the arch. Give your answer to 3 significant figures, using unrounded values.
(3)
(e)
The arch is made as a slab of constant thickness \(0.5\) m whose face is this region. Find the volume of the slab. Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 right-angled triangle \(OMA\) with \(OM = r - 6\) and \(AM = 20\): \(r^2 = 20^2 + (r-6)^2\)
M1 subtract triangle area \(\frac{1}{2}r^2\sin\theta\)
✓ 1
A1 \(163\) m\(^2\)
✓ 1
Part (e)
M1 volume \(= 0.5 \times 162.8\ldots\)
✓ 1
A1 \(81.4\) m\(^3\)
✓ 1
Exam tip: In a circular segment problem, split the figure into a sector and a triangle, and find the central angle first.
309IBMA3S-3-18r40-123.5 The unit circle and the ambiguous casePaper 2Hard14 marks
Calculator
A space probe \(P\) orbits a planet of radius \(3400\) km, at a constant height of \(1200\) km above the surface. Let \(C\) be the centre of the planet, so \(CP = 4600\) km. The line from \(P\) to the horizon touches the surface at a point \(T\).
Diagram not to scale
(a)
Find the distance \(PT\) to the nearest kilometre.
(3)
(b)
Find the angle \(PCT\) in radians, to 3 significant figures.
(2)
(c)
Find the distance along the surface from the point directly below \(P\) to \(T\), to the nearest kilometre, using unrounded values.
(2)
(d)
A second probe \(Q\) orbits at the same height, with angle \(PCQ = 50^\circ\). Find the distance \(PQ\) to the nearest kilometre.
(3)
(e)
Determine whether the straight line from \(P\) to \(Q\) passes through the planet. Justify your answer with a calculation.
(4)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(CT \perp PT\) (tangent perpendicular to radius), so triangle \(CTP\) is right-angled
✓ 1
M1 \(PT^2 = 4600^2 - 3400^2\)
✓ 1
A1 \(PT = 3098\) km
✓ 1
Part (b)
M1 \(\cos(PCT) = \frac{3400}{4600}\)
✓ 1
A1 \(0.739\) rad
✓ 1
Part (c)
M1 arc length \(= 3400 \times 0.739\ldots\)
✓ 1
A1 \(2513\) km
✓ 1
Part (d)
M1 cosine rule in triangle \(PCQ\) (or isosceles triangle split into two right-angled triangles)
M1 let \(N\) be the midpoint of \(PQ\); \(CN \perp PQ\) and angle \(PCN = 25^\circ\)
✓ 1
M1 \(CN = 4600\cos 25^\circ\)
✓ 1
A1 \(CN = 4169\) km
✓ 1
R1 \(4169 > 3400\), so the line does not pass through the planet and the signal is not blocked
✓ 1
Exam tip: For a line joining two points on a circle of radius R, the shortest distance from the centre to the line is found from the right-angled triangle formed with the midpoint.
310IBMA3S-3-17r48-103.3 Applications of trigonometryPaper 2Hard9 marks
Calculator
A rock tower stands vertically on level ground with its top P at a height of 480 m above its base F. Two mountaineers stand at points A and B on the ground. The angle of elevation of P from A is 25° and from B is 32°. The angle AFB is 110°.
Diagram not to scale
(a)
Find FA and FB, giving each to the nearest metre.
(4)
(b)
Find the distance AB between the mountaineers, to the nearest metre. Use unrounded values of FA and FB.
(3)
(c)
Find the area of triangle AFB, in m² to 3 significant figures. Use unrounded values of FA and FB.
M1 attempt to use \(\frac{1}{2}(FA)(FB)\sin 110°\)
✓ 1
A1 area = 3.72 × 105 m²
✓ 1
Common error: The angle of elevation is measured from the horizontal at A or B; draw the vertical right-angled triangle containing P and F first.
311IBMA3S-3-17r48-113.8 Trigonometric equationsPaper 2Hard8 marks
Calculator
The strength \(S\) of a signal in a waveguide is modelled by \(S = 2\cos 2x + 5\sin x\), where \(x\) is in radians and \(0 \le x \le 2\pi\).
(a)
Show that \(S = 3\) can be written as \(4\sin^2 x - 5\sin x + 1 = 0\).
(3)
(b)
Hence find all values of \(x\) for which \(S = 3\), giving non-exact answers to 3 significant figures.
(5)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use \(\cos 2x = 1 - 2\sin^2 x\)
✓ 1
A1 \(2 - 4\sin^2 x + 5\sin x = 3\)
✓ 1
A1 rearrange to \(4\sin^2 x - 5\sin x + 1 = 0\) (AG)
✓ 1
Part (b)
M1 attempt to solve: \((4\sin x - 1)(\sin x - 1) = 0\)
✓ 1
A1 \(\sin x = 1\) or \(\sin x = \frac{1}{4}\)
✓ 1
A1 \(x = \frac{\pi}{2}\) (= 1.57)
✓ 1
A1 \(x = 0.253\)
✓ 1
A1 \(x = 2.89\)
✓ 1
Exam tip: When an equation mixes \(\cos 2x\) and \(\sin x\), rewrite \(\cos 2x\) in terms of \(\sin x\) so that a single trigonometric function remains.
We use essential cookies to make this site work. With your permission we'd also use analytics and advertising cookies to understand which resources help students most and how families find us. Nothing optional is set unless you accept. Privacy policy
EssentialNeeded for the site and your cookie choice to work.
Always on
AnalyticsAnonymous visit statistics that help improve the free resources.
MarketingMeasures whether ads reach the right families. Off unless you turn it on.