Trigonometry & Inverse Trigonometry
Angles of Elevation and Depression
Grade 11
Question:
<p>An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60° and after 10 seconds the elevation is observed to be 30°. The uniform speed of the aeroplane in km/h is</p>
<p>(a) 240</p>
<p>(b) 240√3</p>
<p>(c) 60√3</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Use cotangent relationships in angles of elevation and distance = speed × time to find the uniform speed.
<p><strong>Solution:</strong></p><p>Let the height of the aeroplane be H = 1 km.</p><p>At the first observation (angle of elevation = 60°):</p><p>$d_1 = H \cot 60° = 1 \cdot \frac{1}{\sqrt{3}}$</p><p>At the second observation (angle of elevation = 30°):</p><p>$d_2 = H \cot 30° = 1 \cdot \sqrt{3}$</p><p>Distance travelled by aeroplane:</p><p>$d = d_2 - d_1 = \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{3-1}{\sqrt{3}} = \frac{2}{\sqrt{3}}$ km</p><p>Time taken = 10 seconds = $\frac{10}{3600}$ hours = $\frac{1}{360}$ hours</p><p>Speed = $\frac{d}{t} = \frac{\cot 30° - \cot 60°}{10} \times 3600 = 240\sqrt{3}$ km/h</p><p>∴ Answer is (b) 240√3</p>
Correct Answer: b