The sine rule Quiz (Grade 7)

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Score: 0 / 10

The sine rule — Notes & Worked Examples Show / Hide

Explanation:

Use the Sine Rule when a non-right-angled triangle involves two opposite side-angle pairs. To find a missing side, put sides on top: asin A = bsin B. To find a missing angle, flip the fractions to put sines on top: sin Aa = sin Bb.

Always label each angle with a capital letter (A, B, C) and the side directly opposite it with the matching lowercase letter (a, b, c). Watch out for the ambiguous case if asked for an obtuse angle (180° − θ).

Example 1: In ΔABC, A = 42°, B = 75°, and b = 13.4 cm. Find side a (3 s.f.).

Solution:

Use asin A = bsin B:

asin 42° = 13.4sin 75°

a = 13.4 × sin 42°sin 75° = 9.2827... cm

Answer: a = 9.28 cm (3 s.f.)

Example 2: In ΔPQR, p = 8.2 cm, q = 11.5 cm, and Q = 68°. Find acute angle P (1 d.p.).

Solution:

Use sin Pp = sin Qq ⇒ sin P8.2 = sin 68°11.5

sin P = 8.2 × sin 68°11.5 = 0.66112...

P = sin⁻¹(0.66112...) = 41.385...°

Answer: P = 41.4° (1 d.p.)

One Page Guide (PDF) The sine rule Worksheets

Question 1 AQA Jun 2025 3H Q17

[2 marks]

Triangle with angles 119° and 31°, side 6.3 cm opposite 31°. Use the sine rule to work out x. [2 marks]

Use the sine rule to work out x.

x = cm

Worked Solution & Answer

x / sin(119°) = 6.3 / sin(31°) M1
⇒ x = [6.3 × sin(119°)] / sin(31°) ≈ 10.7 cm. A1

Answer: x = 10.7 cm

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Question 2 AQA Nov 2022 2H Q28

[2 marks]

ABCDEF is a triangular prism.
P is a point on EF.

EB = 29 cm
Angle EBP = 35°
Angle EPB = 114°
Work out the length of EP.

Answer cm

Worked Solution & Answer

By Sine Rule: EP / sin 35° = 29 / sin 114° M1
EP = 29 × sin 35° / sin 114° = 18.21... = 18.2 cm. A1

Answer: Answer 18.2 cm

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Question 3 AQA Nov 2024 3H Q19

[3 marks]

Triangle with angles 35° and 117°, side 54 cm opposite 117°. Use the sine rule to work out x. [3 marks]

Use the sine rule to work out length x.
You **must** show your working.

x = cm

Worked Solution & Answer

x / sin 35° = 54 / sin 117°. M2
x = (54 × sin 35°) ÷ sin 117° ≈ 34.8 cm. A1

Answer: x = 34.8 cm

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Question 4 OCR Jun 2025 4H Q20

[3 marks]

The diagram shows triangle ABC.

AC = 12 cm, angle CAB = 40° and angle ABC = 58°.
Calculate length BC.

Worked Solution & Answer

By the sine rule: BC / sin 40° = 12 / sin 58° M2
BC = (12 × sin 40°) ÷ sin 58° = 7.71345... ÷ 0.848048... = 9.0955... ≈ 9.09 cm (or 9.1 cm). A1

Answer: 9.09 cm

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Question 5 IGCSE Nov 2021 1H Q13

[3 marks]

Here is a triangle XYZ.

The length XZ and the angles YXZ and XYZ are each given correct to 2 significant figures.
Calculate the upper bound for the length YZ.
Give your answer correct to one decimal place.
Show your working clearly.

Worked Solution & Answer

UB(XZ) = 16.5 cm, LB(∠YXZ) = 125° (which maximises sin ∠YXZ on [125°, 135°]), LB(∠XYZ) = 24.5° M2
By Sine Rule: UB(YZ) = (16.5 × sin 125°) / sin 24.5° = 13.516... / 0.41469... = 32.593... = 32.6 cm A1

Answer: 32.6 cm

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Question 6

[4 marks]

ABC is a triangle.

Calculate the size of angle BAC.
Give your answer correct to 1 decimal place.

Worked Solution & Answer

(Angles in a triangle add up to 180°)
sin(BCA) / 15 = sin 70° / 18 ⇒ sin(BCA) = 0.78307... ⇒ ∠BCA = 51.54° M3
∠BAC = 180 – (70 + 51.54) = 58.5° A1

Answer: 58.5

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Question 7

[4 marks]

Here is a triangle ABC.

Work out the value of \sin ABC
Give your answer in the form mn where m and n are integers.

Worked Solution & Answer

sin ABC / 6.5 = sin 30° / 10.7 M3
⇒ sin ABC = (6.5 × 0.5) / 10.7 = 3.25 / 10.7 = 65 / 214 A1

Answer: 65214

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Question 8

[4 marks]

PQR and QRS are triangles.

Calculate the length of QS.
Give your answer correct to 3 significant figures.
You must show all your working.

Worked Solution & Answer

By cosine rule in ΔPQR: QR² = 11² + 9.4² – 2(11)(9.4)cos 27° = 209.36 – 206.8 cos 27° = 25.099... ⇒ QR = 5.0099... cm M3
By sine rule in ΔQRS: QS / sin 88° = QR / sin 41° ⇒ QS = (5.0099 × sin 88°) / sin 41° = 7.631... → 7.63 cm A1

Answer: 7.63 cm

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Question 9 AQA Nov 2023 3H Q24

[4 marks]

Work out the area of triangle ABC.

Answer cm²

Worked Solution & Answer

Angle A = 180° – (73° + 56°) = 51° M2
By the sine rule: BC / sin 51° = 24 / sin 73° ⇒ BC = (24 × sin 51°) / sin 73° = 19.5035... cm M1
Area = 1/2 × 24 × 19.5035... × sin 56° = 193.53... ≈ 193.5 cm². A1

Answer: Answer 193.5 cm²

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Question 10 AQA Jun 2024 3H Q18

[4 marks]

(a) Triangle A has sides 11 cm, 7 cm and angle 35° opposite 7 cm. Use the sine rule to show x = 64° to the nearest degree. [3 marks]
18 (b) Triangle B has 35° between the 11 cm and 7 cm sides. Without further calculation, is w = 64°? [1 mark]

(a) Use the sine rule to show that x = 64° to the nearest degree. [3 marks]

(b) Anna thinks that w must be 64° to the nearest degree.
She says,
“This is because triangle B has two sides and one angle the same as triangle A.”
**Without further calculation**, is she correct?
Tick a box.
Yes [ ] No [ ]
Give a reason for your answer. [1 mark]

Worked Solution & Answer

(a)

sin x / 11 = sin 35° / 7 ⇒ sin x = (11 × sin 35°) / 7 = 0.90134... M2
x = sin⁻¹(0.90134...) = 64.33° = 64° (to nearest degree) C1

Answer (a): sin x / 11 = sin 35° / 7 ⇒ sin x = 11 sin 35° / 7 = 0.9013... ⇒ x = 64.3° ≈ 64°

(b)

No, because in triangle B the 35° angle is the included angle between the 11 cm and 7 cm sides (SAS), whereas in triangle A it is opposite the 7 cm side C1

Answer (b): No — the 35° angle is in a different position (included between 11 cm and 7 cm in B, not opposite the 7 cm side)

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Score: 0 / 10 (0 / 33 marks)

The sine rule Worksheets Next: The cosine rule →

Grade 7 Quizzes

19 Topics

7A - Surds	7B - Bounds
7C - Proportion (harder)	7D - Algebraic fractions
7E - The quadratic formula	7F - Factorising harder quadratics
7G - Trigonometric & exp. Graphs	7H - Rearranging (harder)
7J - Functions (harder)	7K - Iteration
7L - Area of any triangle	7M - The sine rule
7N - The cosine rule	7P - Congruent triangles
7Q - 3D pythagoras / trig.	7R - Histograms
7S - Conditional probability	7T - Exact trig. values (harder)
7U - Area of a segment