Question
The angle of elevation of an aeroplane from a point on
the ground is 60°. After 20 seconds flight the elevation changes to 30°, if the aeroplane is flying at a height of 3000√3 m. find the speed of the aeroplane.Solution
 Distance between two place in which angle change is  D = n(cotθβ + cotθβ) Distance = 3000√3 (cot30° + cot60°) = 3000√3 × √3 - 1/√3  = 3000√3 × 2/√3 = 6000 m So, distance covered by plane in 20 sec = 6000 m Speed of plane = 6000/20 = 300 m/sec.
 Distance between two place in which angle change is  D = n(cotθβ + cotθβ) Distance = 3000√3 (cot30° + cot60°) = 3000√3 × √3 - 1/√3  = 3000√3 × 2/√3 = 6000 m So, distance covered by plane in 20 sec = 6000 m Speed of plane = 6000/20 = 300 m/sec.
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