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 1440√3 m. find the speed of the aeroplane.Solution
Distance between two place in which angle change is D = n(cotθ₁ + cotθ₂) Distance = 1440√3 (cot30° + cot60°) = 1440√3 × √3 - 1/√3 = 1440√3 × 2/√3 = 2880 m So, distance covered by plane in 12 sec = 2880 m Speed of plane = 2880/12 = 240 m/sec.
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