Question

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 1

Let angle of elevation of an aeroplane is 45°. After 15 seconds angle of elevation is the change to 30°. Let DE be the height of aeroplane which is 3000 meters above the ground.

Let AB = x,BD = y, ∠CAB = 45° and ∠EAD = 30°

Here we have to find speed of aero plane.

We have the corresponding figure as follows

So we use trigonometric ratios.

In Δ ABC

`=> tan A = (BC)/(AB)`

`=> tan 45^@ = 3000/x`

`=> 1 = 3000/x`

`=> x = 3000`

Again in  ΔADE

`=> tan A = (DE)/(AB + BD)`

`=> tan 30^@ = 3000/(x + y)`

`=> 1/sqrt3 = 3000/(3000 + y)`

`=> 3000 + y = 3000sqrt3`

`=> y = 3000sqrt3 - 3000`

`=> y = 3000(sqrt3 - 1)`

=> y = 2196

Since 15 sec = 2196

`=> sec = 2196/15 = 146.4`

`= (146.4 xx 3600)/1000`

= 527.04

Hence the speed of aero plane is 527.04 km/h