Trigonometry & Inverse Trigonometry
Heights and Distances
Grade 11
Question:
<p>From point \( D \), 40 m away from the base \( A \) of a vertical tower \( BC \) of height \( h \), the angle of elevation of the top \( C \) is \( 30^\circ \). From point \( B \) (at the base of the tower), the angle of elevation of \( C \) is \( 60^\circ \), and \( B \) is at a horizontal distance \( x \) from \( A \). Find \( x \) (in metres).</p>
<p>10 m</p>
<p>20 m</p>
<p>30 m</p>
<p>40 m</p>
Step-by-Step Solution
Key Concept: Use tan(angle) = opposite/adjacent in two right triangles formed by the tower. The angle of elevation from D (40m from A) is 30°, giving tan(30°) = h/40. Since the angle from B is 60°, use tan(60°) = h/x to find x by eliminating h.
<p><strong>Step 1:</strong> From point D, which is 40 m horizontally from A (base of tower), angle of elevation to C is 30°.</p><p>In right triangle ADC: tan(30°) = h/40</p><p>Therefore: h = 40 · tan(30°) = 40 · (1/√3) = 40/√3 m</p><p><strong>Step 2:</strong> From point B at the base of the tower (horizontal distance x from A), angle of elevation to C is 60°.</p><p>In right triangle ABC: tan(60°) = h/x</p><p>Therefore: √3 = h/x, which gives h = x√3</p><p><strong>Step 3:</strong> Equate the two expressions for h:</p><p>x√3 = 40/√3</p><p>x = 40/(√3 · √3) = 40/3 m</p><p><strong>Step 4:</strong> Rationalize if needed: x = 40/3 ≈ 13.33 m</p><p>∴ Answer: <strong>x = 40/3 metres (or 13⅓ m)</strong></p>
Correct Answer: B