Sine rule, cosine rule, area = ½ab sin C – GCSE Maths exam questions with mark scheme

1Grade 6–7HigherCalculator3 marks

In triangle ABC, angle BAC = 48°, angle ABC = 71° and BC = 9.4 cm.

Work out the length of AC. Give your answer correct to 3 significant figures. [3]

$\dfrac{AC}{\sin71^\circ}=\dfrac{9.4}{\sin48^\circ}$M1
$AC=\dfrac{9.4\sin71^\circ}{\sin48^\circ}$M1
12.0 cm (accept 11.95 to 11.96)A1
Answer: 12.0 cm
Examiner tip: Pair each side with the angle opposite it – BC is opposite A, and AC is opposite B.

2Grade 6–7HigherCalculator3 marks

In triangle ABC, AC = 7 cm, AB = 11 cm and angle BAC = 58°.

Work out the length $x$ of BC. Give your answer correct to 3 significant figures. [3]

$x^2=7^2+11^2-2\times7\times11\times\cos58^\circ$M1
$x^2=88.39\ldots$ and $x=\sqrt{88.39\ldots}$M1
9.40 cm (accept 9.401 to 9.402)A1
Answer: 9.40 cm
Examiner tip: Work out $2bc\cos A$ before subtracting – typing $170-154$ first and then multiplying by cos 58° is a classic calculator error.

3Grade 6–7HigherCalculator2 marks

Triangle PQR has PQ = 8 cm, PR = 13 cm and angle QPR = 115°.

Work out the area of triangle PQR. Give your answer correct to 3 significant figures. [2]

$\tfrac12\times8\times13\times\sin115^\circ$M1
47.1 cm$^2$ (accept 47.12 to 47.13)A1
Answer: 47.1 cm$^2$
Examiner tip: The angle must be the one between the two given sides (the included angle).

4Grade 6–7HigherCalculator3 marks

A triangle has sides of length 5 cm, 7 cm and 9 cm.

Work out the size of the largest angle in the triangle. Give your answer correct to 1 decimal place. [3]

uses the cosine rule with the angle $\theta$ opposite the 9 cm side: $9^2=5^2+7^2-2\times5\times7\times\cos\theta$M1
$\cos\theta=\dfrac{5^2+7^2-9^2}{2\times5\times7}\;(=-0.1)$M1
95.7° (accept 95.73 to 95.74)A1
Answer: 95.7°
Examiner tip: A negative cosine is fine – it means the angle is obtuse; don't drop the minus sign.

5Grade 8–9HigherCalculator4 marks

A boat sails 12 km from port A on a bearing of 050° to a buoy B. It then sails 8 km on a bearing of 165° to a lighthouse C.

(a) Show that angle ABC = 65°. [1]

(b) Work out the distance AC. Give your answer correct to 3 significant figures. [3]

(a) bearing of A from B is $050^\circ+180^\circ=230^\circ$, so angle ABC $=230^\circ-165^\circ=65^\circ$ (or: angle between BA and the south line at B is 50° (alternate angles) and between the south line and BC is $180^\circ-165^\circ=15^\circ$, so $50^\circ+15^\circ=65^\circ$)B1
(b) $AC^2=12^2+8^2-2\times12\times8\times\cos65^\circ$M1
$AC^2=126.85\ldots$ and square rootsM1
11.3 km (accept 11.26)A1
Answer: (a) shown (b) 11.3 km
Examiner tip: Draw a north line at every point – angles at B come from the back bearing (230°), not from the original 050°.

6Grade 8–9HigherCalculator5 marks

Triangle XYZ has XY = 9 cm, angle YXZ = 40° and area 30 cm$^2$.

Work out the length of YZ. Give your answer correct to 3 significant figures. [5]

$\tfrac12\times9\times XZ\times\sin40^\circ=30$M1
$XZ=10.37\ldots$A1
$YZ^2=9^2+10.37^2-2\times9\times10.37\times\cos40^\circ$M1
$YZ^2=45.55\ldots$ and square rootsM1
6.75 cm (accept 6.74 to 6.75)A1
Answer: 6.75 cm
Examiner tip: Use the area formula backwards to find XZ first, then keep it unrounded for the cosine rule.

7Grade 8–9HigherCalculator5 marks

ABCD is a quadrilateral. AB = 6 cm, AD = 8 cm and angle BAD = 72°. Angle BCD = 45° and angle CBD = 60°.

Work out the length of CD. Give your answer correct to 3 significant figures. [5]

$BD^2=6^2+8^2-2\times6\times8\times\cos72^\circ$M1
$BD=8.386\ldots$ ($BD^2=70.33\ldots$)A1
$\dfrac{CD}{\sin60^\circ}=\dfrac{BD}{\sin45^\circ}$M1
$CD=\dfrac{8.386\ldots\times\sin60^\circ}{\sin45^\circ}$M1
10.3 cm (accept 10.27)A1
Answer: 10.3 cm
Examiner tip: Spot the shared diagonal BD – find it with the cosine rule in one triangle, then use it in the sine rule in the other.

8Grade 6–7HigherCalculator2 marks

The diagram shows a triangular sail ABC.

AB = 5.2 m, AC = 3.6 m and angle BAC = 64°.

Work out the area of the sail. Give your answer correct to 3 significant figures. [2]

$\frac{1}{2}\times5.2\times3.6\times\sin64^\circ$M1
8.41 m$^2$ (accept 8.412…; units not required for the mark)A1
Answer: 8.41 m$^2$
Examiner tip: Use the two sides that meet at the given angle – here 5.2 and 3.6 meet at A, so no other side is needed; check your calculator is in degrees.

9Grade 6–7HigherNon-calculator5 marks

Triangle PQR has PQ = 5 cm, PR = 8 cm and angle QPR = 60°.

(a) Show that QR = 7 cm. [3]

(b) Work out the area of triangle PQR. Give your answer in the form $a\sqrt{3}$ cm$^2$, where $a$ is an integer. [2]

(a) $QR^2=5^2+8^2-2\times5\times8\times\cos60^\circ$M1
uses $\cos60^\circ=\frac{1}{2}$, e.g. $25+64-40$M1
$QR^2=49$ so $QR=\sqrt{49}=7$ cm, with no errors seen (answer given, so all working must be shown)A1
(b) $\frac{1}{2}\times5\times8\times\frac{\sqrt{3}}{2}$ (uses $\sin60^\circ=\frac{\sqrt3}{2}$)M1
$10\sqrt{3}$ (cm$^2$)A1
Answer: (a) shown (b) $10\sqrt{3}$ cm$^2$
Examiner tip: On the non-calculator paper you must know $\cos60^\circ=\frac12$ and $\sin60^\circ=\frac{\sqrt3}{2}$ – and for a ‘show that’ the line $QR^2=49$ must appear before $QR=7$.

10Grade 6–7HigherCalculator3 marks

In triangle ABC, AB = 9 cm, AC = 6 cm and angle BAC = 50°. Ravi works out the length $x$ of BC. His working is shown.

$x^2 = 9^2+6^2-2\times9\times6\times\cos50^\circ$
$x^2 = 117-108\times\cos50^\circ$
$x^2 = 9\times\cos50^\circ$
$x^2 = 5.785\ldots$
$x = 2.41$ cm

(a) Describe the mistake Ravi has made. [1]

(b) Work out the correct length of BC. Give your answer correct to 3 significant figures. [2]

(a) he subtracted 108 from 117 before multiplying by $\cos50^\circ$ (he worked out $(117-108)\cos50^\circ$); he should have worked out $108\cos50^\circ$ first and then subtracted it from 117. Accept any clear statement that the order of operations (multiplication before subtraction) is wrong. Do not accept just ‘he used the wrong formula’C1
(b) $117-69.4\ldots$ or $x^2=47.57\ldots$ (accept 47.6)M1
6.90 cm (accept 6.897…; accept 6.9)A1
Answer: (a) He did $117-108$ first instead of multiplying $108\times\cos50^\circ$ first. (b) 6.90 cm
Examiner tip: Type the whole expression $117-108\cos50$ into your calculator in one go – working left to right breaks BIDMAS and costs every mark after that line.

11Grade 6–7HigherCalculator3 marks

In triangle DEF, DE = 8.3 cm, EF = 11.6 cm and angle EDF = 74°.

Angle DFE = $\theta$.

Work out the size of angle $\theta$. Give your answer correct to 1 decimal place. [3]

$\dfrac{\sin\theta}{8.3}=\dfrac{\sin74^\circ}{11.6}$ oe (correct sine rule with the sides opposite the angles)M1
$\sin\theta=\dfrac{8.3\times\sin74^\circ}{11.6}$ $(=0.6877\ldots)$M1
43.5° (accept 43.45…)A1
Answer: 43.5°
Examiner tip: Pair each angle with the side opposite it: 74° is opposite EF (11.6 cm) and $\theta$ is opposite DE (8.3 cm) – then use $\sin^{-1}$ at the end.

12Grade 6–7HigherCalculator4 marks

A triangle has two sides of length 7.5 cm and 12 cm. The area of the triangle is 36 cm$^2$.

(a) Given that the angle between these two sides is acute, work out the size of this angle. Give your answer correct to 1 decimal place. [3]

(b) Lena says, ‘There is only one possible size for the angle between the two sides.’ Explain why Lena is wrong. [1]

(a) $\frac{1}{2}\times7.5\times12\times\sin C=36$M1
$\sin C=\dfrac{36}{45}$ $(=0.8)$M1
53.1° (accept 53.13…)A1
(b) the angle could be obtuse: $180^\circ-53.1^\circ=126.9^\circ$ has the same sine ($0.8$), so it gives the same area. Accept a statement that $\sin(180^\circ-C)=\sin C$ or a correct value 126.9° (accept 127°) with reasonC1
Answer: (a) 53.1° (b) The angle could also be $126.9^\circ$, since $\sin126.9^\circ=\sin53.1^\circ=0.8$.
Examiner tip: Rearrange before using $\sin^{-1}$: divide 36 by $\frac12\times7.5\times12=45$ – then remember the calculator only gives the acute answer.

13Grade 8–9HigherCalculator4 marks

The diagram shows a circle with centre O and radius 9 cm. A and B are points on the circle and angle AOB = 110°.

Work out the area of the shaded segment. Give your answer correct to 3 significant figures. [4]

area of sector: $\dfrac{110}{360}\times\pi\times9^2$ $(=77.75\ldots)$M1
area of triangle AOB: $\frac{1}{2}\times9\times9\times\sin110^\circ$ $(=38.05\ldots)$M1
sector − triangle, e.g. $77.75\ldots-38.05\ldots$M1
39.7 cm$^2$ (accept answers in the range 39.6 to 39.7 from premature rounding)A1
Answer: 39.7 cm$^2$
Examiner tip: The triangle is not right-angled, so ‘½ × base × height’ with 9 and 9 is wrong – use $\frac12 ab\sin C$ with the 110° angle.

14Grade 8–9HigherCalculator6 marks

The diagram shows a triangular field PQR. PQ = 120 m, PR = 85 m and QR = 140 m.

A farmer wants to sow grass seed over the whole field. One bag of seed covers 750 m$^2$.

Work out the least number of bags the farmer needs to buy. You must show your working. [6]

$\cos P=\dfrac{85^2+120^2-140^2}{2\times85\times120}$ (or a correct cosine rule for any angle of the triangle)M1
$P=84.3^\circ$ (accept 84.30…; or $Q=37.2^\circ$ or $R=58.5^\circ$ if the cosine rule was used for another angle)A1
$\frac{1}{2}\times85\times120\times\sin84.3^\circ$ (or $\frac12\times$ the two sides that meet at their found angle $\times$ its sine)M1
area $=5070$ m$^2$ (accept 5070 to 5080, e.g. 5074.8…, from correctly rounded angles)A1
their area $\div750$ $(=6.76\ldots)$M1
7 (bags) – must come from a correct area; do not accept 6A1
Answer: 7 bags
Examiner tip: You can't use $\frac12ab\sin C$ until you have an angle, so find one with the cosine rule first and keep it unrounded in your calculator – then round the number of bags UP, because part of a bag can't be bought.

15Grade 8–9HigherCalculator5 marks

A and B are two points 50 m apart on one straight bank of a river. A tree T stands on the opposite bank, which is parallel to the first.

Angle TAB = 58° and angle TBA = 67°.

Work out the width of the river. Give your answer correct to 3 significant figures. [5]

angle ATB $=180^\circ-58^\circ-67^\circ=55^\circ$B1
$\dfrac{AT}{\sin67^\circ}=\dfrac{50}{\sin55^\circ}$ (or $\dfrac{BT}{\sin58^\circ}=\dfrac{50}{\sin55^\circ}$)M1
$AT=56.18\ldots$ (or $BT=51.76\ldots$)A1
width $=AT\times\sin58^\circ$ (or $BT\times\sin67^\circ$) – right-angled trigonometry in the triangle formed by the perpendicular from T. Alternative method: $\dfrac{w}{\tan58^\circ}+\dfrac{w}{\tan67^\circ}=50$ oe, with $w$ the width, earns the first four marks (B1 M1 A1 M1) when fully correctM1
47.6 m (accept answers in the range 47.6 to 47.7, e.g. 47.65 from rounding AT to 56.19 first)A1
Answer: 47.6 m
Examiner tip: AT is not the width – the width is the perpendicular distance from T to the bank, so finish with $\sin58^\circ$ in a right-angled triangle.

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