Sine rule

The method

1. Draw a labelled sketch. Mark vertical and horizontal lines.
2. Use equal alternate angles to connect depression and elevation when the horizontal lines are parallel.
3. For a right triangle, select the ratio using the sides relative to the angle.
4. For the sine rule, pair each side with its opposite angle: a ÷ sin A = b ÷ sin B.
5. Find a missing third angle by subtracting the other two from 180°.
6. When finding an angle with the sine rule, also check its supplement. Keep any answer that gives a valid triangle.

Mistakes that lose marks

• Measuring elevation from the vertical.
• Using the full building height instead of the difference between two tops.
• Pairing a side with an angle beside it instead of opposite it.
• Missing a second valid angle in an ambiguous sine rule problem.

EXAM-STYLE questions written by Apex tutors

Practise with 8 original questions and full solutions. The suggested marks are a guide from our tutors. M means method, A means accuracy and B means an independent point. Try each question, then open its solution.

1. Apex original extended On level ground, a point is 45 m horizontally from a vertical tower. The angle of elevation to its top is 38°. Find the height to 3 significant figures. [3 marks]
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   1. tan 38° = h ÷ 45. M1
   2. h = 45 × tan 38° = 35.157853… m = 35.2 m. A2

   Answer: 35.2 m

2. Apex original extended From the top of a vertical cliff 60 m high, the angle of depression to a boat at sea level is 22°. Find its horizontal distance from the cliff base to 3 significant figures. [3 marks]
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   1. The elevation angle from the boat is also 22° because the horizontal lines are parallel. M1
   2. tan 22° = 60 ÷ d. M1
   3. d = 60 ÷ tan 22° = 148.505211… m = 149 m. A1

   Answer: 149 m

3. Apex original extended A vertical flagpole is 8 m tall. A point on level ground is 15 m from its base. Find the elevation angle from that point to its top to 1 decimal place. [3 marks]
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   1. tan θ = 8 ÷ 15. M1
   2. θ = tan⁻¹(8 ÷ 15) = 28.072487…° = 28.1°. A2

   Answer: 28.1°

4. Apex original extended Building A is 3.4 m high and building B is 1.6 m high. They stand on level ground, 4.0 m apart horizontally. Find the height difference and the angle of depression from the top of A to the top of B to 1 decimal place. [5 marks]
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   1. Height difference = 3.4 − 1.6 = 1.8 m. M1
   2. tan θ = 1.8 ÷ 4.0 = 0.45. M1
   3. θ = tan⁻¹(0.45) = 24.227745…° = 24.2°. A3

   Answer: 1.8 m; 24.2°

5. Apex original extended In triangle ABC, A = 52°, B = 61° and side c opposite C is 14 cm. Find side a opposite A to 3 significant figures. [3 marks]
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   1. C = 180° − 52° − 61° = 67°. M1
   2. a ÷ sin 52° = 14 ÷ sin 67°. M1
   3. a = 14 × sin 52° ÷ sin 67° = 11.984891… cm = 12.0 cm. A1

   Answer: 12.0 cm

6. Apex original extended In triangle PQR, P = 40°, Q = 75° and side p opposite P is 9 cm. Find side r opposite R to 3 significant figures. [3 marks]
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   1. R = 180° − 40° − 75° = 65°. M1
   2. r ÷ sin 65° = 9 ÷ sin 40°. M1
   3. r = 9 × sin 65° ÷ sin 40° = 12.689682… cm = 12.7 cm. A1

   Answer: 12.7 cm

7. Apex original extended In triangle XYZ, x = 11 cm is opposite X, y = 8 cm is opposite Y, and Y = 34°. Find all possible values of X to 1 decimal place. [4 marks]
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   1. sin X ÷ 11 = sin 34° ÷ 8. M1
   2. sin X = 11 × sin 34° ÷ 8 = 0.76889024… . M1
   3. First angle = sin⁻¹(11 × sin 34° ÷ 8) = 50.254338…° = 50.3°. M1
   4. Second angle = 180° − 50.254338…° = 129.745662…° = 129.7°.
   5. Remaining angles are 180° − 34° − 50.254338…° = 95.745662…° and 180° − 34° − 129.745662…° = 16.254338…°. Both are positive, so both triangles are valid. A1

   Answer: 50.3° or 129.7°

8. Apex original extended A vertical tower has top T and base B on level ground. B, D and E lie in that order on a straight line. DE = 20 m. Elevation angles to T are 29° at D and 19° at E. Show that angle DTE = 10°. Find DT and the tower height to 3 significant figures. [6 marks]
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   1. Angle TDE = 180° − 29° = 151°. M1
   2. Angle DTE = 180° − 151° − 19° = 10°. M1
   3. DT ÷ sin 19° = 20 ÷ sin 10°. M1
   4. DT = 20 × sin 19° ÷ sin 10° = 37.497446… m = 37.5 m. M1
   5. In right triangle BDT, height = DT × sin 29°. M1
   6. Use unrounded DT: height = (20 × sin 19° ÷ sin 10°) × sin 29° = 18.179122… m = 18.2 m. A1

   Answer: 10°; DT = 37.5 m; height = 18.2 m

Where this is in the syllabus

Explore the matching courses: Cambridge O Level Mathematics (Syllabus D) 4024.

For the official syllabus, use the board links: Cambridge O Level Mathematics (Syllabus D) 4024.

Questions students ask