0580/42 May/Jun 2023, Question 7: Speed, Distance and Time
Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 11 marks
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The question, in short
Using a triangle with AC = 60 km, CB = 87 km, B due east of A and bearing of C from A 038°, find angle ACB and the average speed over the full round trip in 3 h 20 min.
Summarised in our own words. Use the official paper for the exact wording and the diagram.
Part (a) · 5 marks
Show that angle ACB is 95.1° to 1 decimal place using the given lengths and bearing.
- 1Angle CAB = 90 − 38 = 52°
- 2sin B/60 = sin 52/87
- 3sin B = 60 × sin 52 ÷ 87 = 0.5434…, so angle B = 32.91…°
- 4Angle ACB = 180 − 52 − 32.91… = 95.08… = 95.1° (1 d.p.)
AnswerAngle CAB = 52, then 180 − 52 − sin⁻¹((60 sin 52)/87) = 95.08…
Part (b) · 6 marks
Car goes A to C to B then directly back to A in 3 hours 20 minutes; find average speed in km/h.
- 1AB² = 60² + 87² − 2 × 60 × 87 × cos 95.1°
- 2AB² = 12 097.0…, so AB = 109.98… km
- 3Total distance = 60 + 87 + 109.98… = 256.98… km
- 4Time = 3 h 20 min = 3 20/60 = 10/3 hours
- 5Speed = 256.98… ÷ 10/3 = 77.09… ≈ 77.1 km/h
Answer77.1 or 77.08 to 77.11
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Also searched as: 0580/42/M/J/23 · 0580_s23_qp_42 · 0580 s23 42 · question 7.
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