0580/43 May/Jun 2022, Question 7: The Ambiguous Case of the Sine Rule
Cambridge IGCSE Mathematics 0580 · Paper 43 (Extended) · Calculator · 14 marks
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The question, in short
Solve triangle ABC from side lengths 82 m and 55 m with included angle 76°, then apply to a gate position and a second triangle PQR of equal area.
Summarised in our own words. Use the official paper for the exact wording and the diagram.
Part (a) · 3 marks
Find the length of side BC using the two given sides and the 76° angle.
- 1BC² = 82² + 55² − 2 × 82 × 55 × cos 76
- 2BC² = 7566.8…
- 3BC = 86.98… ≈ 87.0 m
Answer87.0 or 86.98 to 86.99
Part (b) · 3 marks
Find angle ACB of the triangle.
- 1sin ACB/82 = sin 76/86.98…
- 2sin ACB = 82 × sin 76 ÷ 86.98… = 0.9147…
- 3Angle ACB = 66.15… ≈ 66.2°
Answer66.1 or 66.2 or 66.13 to 66.17
Part (c) · 3 marks
Point G on AB is closest to C; find the distance AG.
- 1CG is perpendicular to AB, so triangle AGC is right-angled at G
- 2AG = 55 × cos 76
- 3= 13.30… ≈ 13.3 m
Answer13.3 or 13.30 to 13.31
Part (d) · 5 marks
Triangle PQR has the same area as ABC, with PQ = 90 m and QR = 60 m; find both possible values of angle PQR.
- 1Area of ABC = 1/2 × 82 × 55 × sin 76 = 2188.0… m²
- 21/2 × 90 × 60 × sin PQR = 2188.0…
- 3sin PQR = 2188.0… ÷ 2700 = 0.8103…
- 4PQR = 54.13… ≈ 54.1° or 180 − 54.13… = 125.86… ≈ 125.9°
Answer54.1 or 54.13… and 125.9 or 125.86 to 125.87
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