Question

An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°. Find the speed of the aeroplane in km/hr.

Answer 1

An aeroplane is flying 1 km above the ground making an angle of elevation of aeroplane 60°. After 10 seconds angle of elevation is changed to 30°.

Let CE = x, BC = y, ∠AEB = 30°, ∠DEC = 60°, AB = 1 km and CD = 1 km. 

Here we have to find the speed of an aeroplane.

The corresponding figure is as follows

So we use trigonometric ratios.

In ΔDCE

`=> tan 60^@ = 1/x`

`=> sqrt3 = 1/x`

`=> x = 1/sqrt3`

Again in Δ ABE

`=> tan 30^@ = 1/(x + y)`

`=> 1/sqrt3 =    1/(x + y)`

`=> x + y = sqrt3`

`=> y = sqrt3 - 1/sqrt3`

`=> y = 2/sqrt3``

speed = "distance"/"time"`

`= y/(10 sec)`

`= (2/sqrt3)/(10/(60 xx 60))`

= 415.68

Hence the speed of aeroplace is 415.68 km/h.