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An aeroplane at an altitude of 1500 m makes an angle of depression of 60 degrees with a fixed point on the ground. After 15 seconds of horizontal flight, the angle becomes 30 degrees. The speed of the aeroplane is
A200 m/s
B100 m/s
C100 (square root of 3) m/s
D200 (square root of 3) m/s
Answer & Solution
Correct answer: D. 200 (square root of 3) m/s
1. Initial horizontal distance = 1500/tan(60) = 1500/(square root of 3) = 500 (square root of 3).
2. New horizontal distance = 1500/tan(30) = 1500 (square root of 3).
3. Distance covered = 1500 (square root of 3) - 500 (square root of 3) = 1000 (square root of 3); speed = 1000 (square root of 3)/15 ≈ 200 (square root of 3) m/s.
_Source: NCERT Class 10 Mathematics, Ch 9 "Some Applications of Trigonometry", §9.2_
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