0580/23 Oct/Nov 2022, Question 21: Bearings with Sine and Cosine Rules
Cambridge IGCSE Mathematics 0580 · Paper 23 (Extended) · Calculator · 5 marks
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The question, in short
Using triangle ABC with AC = 17.6 km, BC = 12.8 km, angle BAC = 25°, bearing of C from B = 112° and obtuse angle ABC, find the bearing of B from A.
Summarised in our own words. Use the official paper for the exact wording and the diagram.
Solution · 5 marks
- 1sin ABC / 17.6 = sin 25 / 12.8
- 2sin ABC = 17.6 × sin 25 ÷ 12.8 = 0.5811…
- 3Acute angle = 35.53°, but ABC is obtuse: ABC = 180 − 35.53 = 144.47°
- 4Angle between South at B and BC = 180 − 112 = 68°
- 5Bearing of B from A = 144.47 − 68 = 76.47 (alternate angles between North lines) ≈ 076°
Answer076 or 076.4 to 076.5
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Also searched as: 0580/23/O/N/22 · 0580_w22_qp_23 · 0580 w22 23 · question 21.
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