1. In triangle ABC, a = 6, b = 8 and γ = 30°. The area of the triangle is
- A24
- B12
- C12√3
- D48
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Answer: B. Δ = ½ab sin γ = ½ × 6 × 8 × ½ = 12. Using cos 30° instead of sin 30° gives 12√3.
By Team Megalecture · Updated 4 Nov 2026
20 exam-style questions on Application of trigonometry, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.
20 questions · 2 topics · 10 per test
Try each one, then open the answer. These ten stay the same; the test draws new ones every time.
Answer: B. Δ = ½ab sin γ = ½ × 6 × 8 × ½ = 12. Using cos 30° instead of sin 30° gives 12√3.
Answer: A. The angle in the formula is the one between the two sides b and c, opposite to a. When α = 90° it reduces to Pythagoras.
Answer: C. Δ = ½ab sin 60° = ½ × 4 × 4 × √3/2 = 4√3 cm2.
Answer: A. s = 15, Δ = √(15 × 10 × 3 × 2) = √900 = 30. It is right-angled, so ½ × 5 × 12 = 30 agrees.
Answer: B. s = 6, so sin(γ/2) = √((3)(2)/(12)) = √(1/2) = 1/√2. Then γ/2 = 45° and γ = 90°, as expected for a 3-4-5 triangle.
Answer: C. tan 30° = h/50, so h = 50/√3 ≈ 28.9 m. Using tan 60° gives 86.6 m; 25 m and 43.3 m use sin and cos wrongly.
Answer: A. (a − b)/(a + b) = tan((α − β)/2)/tan((α + β)/2): difference over sum on both sides, using half-angles.
Answer: C. Δ = 6 and s = 6, so r = Δ/s = 1. 2.5 is the circum-radius and 6 is the area.
Answer: C. From the law of sines, a/sin α = 2R, so R = a/(2 sin α). Equivalently R = abc/(4Δ), not abc/Δ.
Answer: B. r = Δ/s = 84/21 = 4, where s is the semi-perimeter. Dividing by the full perimeter 42 gives 2.
20, each with the answer and a short explanation, across 2 topics: Solving triangles, Area of a triangle and connected circles. Every chapter test draws ten of them, so you can take it many times.
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The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.
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