Question
The angle of elevation of an aeroplane from a point on
the ground is 60°. After 15 seconds flight the elevation changes to 30°, if the aeroplane is flying at a height of 2400√3 m. find the speed of the aeroplane.Solution
Distance between two place in which angle change is D = n(cotθ₁ + cotθ₂) Distance = 2400√3 (cot30° + cot60°) = 2400√3 × √3 - 1/√3 = 2400√3 × 2/√3 = 4800 m So, distance covered by plane in 15 sec = 4800 m Speed of plane = 4800/15 = 320 m/sec.
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