Question
The angle of elevation of an aeroplane from a point on
the ground is 60°. After 12 seconds flight the elevation changes to 30°, if the aeroplane is flying at a height of 1200√3 m. find the speed of the aeroplane.Solution
Distance between two place in which angle change is D = n(cotθβ + cotθβ) Distance = 1200√3 (cot30° + cot60°) = 1200√3 × √3 - 1/√3 = 1200√3 ×2/√3 = 2400 m So, distance covered by plane in 12 sec = 2400 m Speed of plane = 2400/12 = 200 m/sec.
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