Trigonometry & Inverse Trigonometry
Heights and Distances
Grade 11
Question:
<p>A tower stands at the centre of a circular park. <em>A</em> and <em>B</em> are two points on the boundary of the park such that <em>AB</em> (= <em>a</em>) subtends an angle of 60° at the foot of the tower, and the angle of elevation of the top of the tower from <em>A</em> or <em>B</em> is 30°. The height of the tower is</p>
<p>\(\dfrac{2a}{\sqrt{3}}\)</p>
<p>\(2a\sqrt{3}\)</p>
<p>\(\dfrac{a}{\sqrt{3}}\)</p>
<p>\(a\sqrt{3}\)</p>
Step-by-Step Solution
Key Concept: Use the angle subtended at the center (60°) to find the radius of the circular park via the chord-angle relationship, then apply the angle of elevation (30°) to find height using tan(30°) = h/r.
<p><strong>Step 1:</strong> Let O be the foot of the tower (center of park) with radius r. Points A and B lie on the circle with AB = a, and ∠AOB = 60°.</p><p><strong>Step 2:</strong> Using the chord-length formula: a = 2r sin(30°) = 2r · (1/2) = r. Therefore, <strong>r = a</strong>.</p><p><strong>Step 3:</strong> The angle of elevation of the top T from point A is 30°. In right triangle OAT: tan(30°) = h/OA = h/r.</p><p><strong>Step 4:</strong> Since tan(30°) = 1/√3, we have: 1/√3 = h/a, so <strong>h = a/√3 = a√3/3</strong>.</p><p>∴ Answer: <strong>h = a/√3</strong> or <strong>a√3/3</strong> (which is option C)</p>
Correct Answer: C