Trigonometry-17
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The angle of elevation of an aeroplane from P on the ground is 300.
After 15 seconds flight horizontally towards P, the angle of elevation changes to 450.
If the aeroplane is flying at a height of 3000 m, find the speed of the plane.
Solution: Let A' be the first position and A be the second position.
In triangle PA'R, A'R = 3000 m
∠A'PR = 300
Therefore, A'R/PR = tan 300
PR = A'R / tan 300 = 3000 / 1/√3 = 3000√3 m
In triangle PAS, AS = 3000 m
∠APS = 450
AS / PS = tan 450
PS = AS / tan 450 = 3000/1 = 3000 m
Distance traveled by the plane in 15 seconds = A'A = RS
PR - PS = 3000√3 - 3000 = 3000 X 0.732 m = 2196 m
Therefore, Speed of the plane = (2196 X 3600) / (15 X 1000) km/h
= 527 km/hr (approximately)
