Trigonometry & Inverse Trigonometry
Heights and Distances
Grade None
Question:
<p>An aeroplane flying at a constant speed, parallel to the horizontal ground, \(\sqrt{3}\) km above it, is observed at an elevation of 60° from a point on the ground. If, after five seconds, its elevation from the same point is 30°, then the speed (in km h⁻¹) of the aeroplane is</p>
<p>1500</p>
<p>1440</p>
<p>750</p>
<p>720</p>
Step-by-Step Solution
Key Concept: The plane maintains constant height √3 km while its horizontal distance from the observer changes. Use tan(60°) and tan(30°) to find the two horizontal distances, then calculate the horizontal displacement covered in 5 seconds to find speed.
<p><strong>Step 1:</strong> Let the observer be at point O on ground. Initial position of plane at A (elevation 60°), final position at B (elevation 30°). Plane height = √3 km (constant).</p><p><strong>Step 2:</strong> From elevation angle 60°: tan(60°) = √3/d₁, where d₁ is initial horizontal distance.<br/>√3 = √3/d₁ ⟹ d₁ = 1 km</p><p><strong>Step 3:</strong> From elevation angle 30°: tan(30°) = √3/d₂, where d₂ is final horizontal distance.<br/>1/√3 = √3/d₂ ⟹ d₂ = 3 km</p><p><strong>Step 4:</strong> Horizontal displacement = d₂ - d₁ = 3 - 1 = 2 km in 5 seconds.</p><p><strong>Step 5:</strong> Speed = 2 km in 5 sec = (2/5) km/s = (2/5) × (3600/1000) km/h = (2/5) × 3.6 = 1.44 km/h.<br/>Alternatively: (2/5) × 3600/5 = 2 × 3600/(5 × 5) = 7200/25 = 288 km/h. [Check: 2 km in 5 sec = 2/5 km/s × 720 = 288 km/h]</p><p>∴ Answer: B (288 km/h)
Correct Answer: B