Q. 444.1( 14 Votes )

The angle of elevation of an aeroplane from a point on the ground is 45°. After a flight of 15 seconds, the elevation changes to 30°. If the aeroplane is flying at a height of 3000 metres, find the speed of the aeroplane.

Answer :

In the fig let C be the initial position of the aeroplane. After 15 seconds the position of the aeroplane becomes E.


tan 45° =

1 =

Y = 3000m ------(2)


tan 30° =


x + y = 3000

Using equation (1) to replace value of y, we get

x = 3000√3 – 3000



Since the distance travelled by aeroplane in 15 seconds is 2196m. Therefore distance travelled by aeroplane in 1 hour =

527.04 km/hr

Therefore speed of aeroplane is 527.04 km/hr

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