A triangle ABC has sides AB = (2x − 1) cm and AC = (2x + 1) cm with angle BAC = 30°, and area (x² + x − 3.75) cm². Determine the largest angle, to the nearest degree.

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Area of triangle = ½ × AB × AC × sin A ½ × (2x − 1)(2x + 1) × sin 30° = x² + x − 3.75 (2x − 1)(2x + 1) = 4x² − 1 and sin 30° = 0.5 0.25 × (4x² − 1) = x² + x − 3.75 x² − 0.25 = x² + x − 3.75 −0.25 = x − 3.75 x = 3.75 − 0.25 = 3.5 AB = 2 × 3.5 − 1 = 6 cm and AC = 2 × 3.5 + 1 = 8 cm, with angle A = 30° BC² = 6² + 8² − 2 × 6 × 8 × cos 30° BC² = 100 − 83.14 = 16.86 BC = √16.86 = 4.106 cm AC = 8 cm is the longest side, so the largest angle is ABC, the angle opposite it cos ABC = (6² + 4.106² − 8²) ÷ (2 × 6 × 4.106) cos ABC = (36 + 16.86 − 64) ÷ 49.28 = −11.14 ÷ 49.28 = −0.226 ABC = 103.06° The largest angle is 103° Answer: 103

4MA1/1HR Jun 2023, Question 25: Choosing Between Sine and Cosine Rules

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