65 exam-style questions on Sine rule, cosine rule; area of triangle — topic 3.2 of Geometry and Trigonometry in the IB Mathematics: Analysis and Approaches syllabus — with answers and explanations. Samples below; the full set is free with an account.
In triangle ABC, angle A = 30°, angle B = 45° and a = 6. Use the sine rule to determine b.
Answer: 6√2
b = a·sin B / sin A = 6 × (√2/2) ÷ (1/2) = 6√2. Dividing by sin B instead of sin A gives 3√2.
A triangle has sides 5, 7 and 8. Determine the size of the angle opposite the side of length 7.
Answer: 60°
cos θ = (5² + 8² − 7²)/(2 × 5 × 8) = 40/80 = 1/2, so θ = 60°.
A triangle has area 15. Two of its sides have lengths 5 and 12, and the included angle θ is acute. Determine θ.
Answer: 30°
15 = (1/2)(5)(12)sin θ = 30 sin θ, so sin θ = 1/2 and the acute solution is θ = 30°.
A triangle has sides 3, 5 and 7. Determine the size of its largest angle.
Answer: 120°
The largest angle is opposite the side of length 7: cos θ = (9 + 25 − 49)/(2 × 3 × 5) = −15/30 = −1/2, so θ = 120°.
Two ships leave a port at the same time. Ship P sails 6 km due north; ship Q sails 8 km on a bearing of 060°. Determine the distance between the ships.
Answer: 2√13 km
The angle between the paths is 60°, so d² = 6² + 8² − 2(6)(8)cos 60° = 100 − 48 = 52, giving d = √52 = 2√13 km. Adding 48 instead of subtracting gives 2√37.
In triangle ABC, a = 5, angle A = 30° and b = 5√2. Determine the acute value of angle B.
Answer: 45°
sin B = b·sin A / a = 5√2 × (1/2) ÷ 5 = √2/2, so the acute solution is B = 45° (the obtuse possibility would be 135°).
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