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0580/42 Feb/Mar 2025, Question 17: Bearings with Sine and Cosine Rules

Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 9 marks

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The question, in short

Using a diagram of points A, B and C on level ground with AC = 65 m, BC = 95 m, angle ACB = 38° and C bearing 145° from A, work out bearings and the length AB.

Summarised in our own words. Use the official paper for the exact wording and the diagram.

Part (a) · 2 marks

Determine the bearing of B as measured from C.

  1. 1Bearing of A from C = 145 + 180 = 325°
  2. 2Bearing of B from C = 325 − 38 = 287°

Answer287

Part (b) · 3 marks

Demonstrate that the distance AB equals 59.3 m to 1 decimal place.

  1. 1AB² = 65² + 95² − 2 × 65 × 95 × cos 38°
  2. 2= 13 250 − 9731.93… = 3518.07…
  3. 3AB = √3518.07… = 59.31… ≈ 59.3 m

Answer√(65² + 95² − 2 × 65 × 95 cos 38) = 59.31…

Part (c) · 4 marks

Given that angle BAC is obtuse, calculate the bearing of B measured from A.

  1. 1sin A/95 = sin 38°/59.3
  2. 2sin A = 95 × sin 38°/59.3 = 0.9863…
  3. 3A is obtuse: angle BAC = 180 − 80.51 = 99.49…°
  4. 4Bearing of B from A = 145 + 99.49…
  5. 5= 244.49… ≈ 244.5°

Answer244.4 to 244.6

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