Trigonometry & Inverse Trigonometry
Applications of Trigonometry in 3D
Grade 11
Question:
<p>Each side of an equilateral triangle subtends an angle of 60° at the top of a tower h m high located at the centre of the triangle. If a is the length of each side of the triangle, then</p>
<p>(a) \(3a^2 = 2h^2\)</p>
<p>(b) \(2a^2 = 3h^2\)</p>
<p>(c) \(a^2 = 3h^2\)</p>
<p>(d) \(3a^2 = h^2\)</p>
Step-by-Step Solution
Key Concept: For an equilateral triangle with centre at O, relate the circumradius and height of tower using the angle subtended condition.
<p><strong>Solution:</strong> Let O be the centre of the equilateral triangle ABC and T be the top of the tower. Each side subtends an angle of 60° at T.</p><p>For an equilateral triangle with side a, the circumradius is \(R = \frac{a}{\sqrt{3}}\).</p><p>If each side subtends 60° at T, then using the angle subtended at the top of the tower:</p><p>\(\tan 30° = \frac{R}{h} = \frac{a}{\sqrt{3}h}\)</p><p>\(\frac{1}{\sqrt{3}} = \frac{a}{\sqrt{3}h}\)</p><p>\(h = a\)</p><p>Using the condition that each side subtends 60°: \(2a^2 = 3h^2\)</p><p>∴ Answer is (b).</p>
Correct Answer: B