The angle of elevation of an aeroplane from point A on the ground is 60˚. After flight of 15 seconds, the angle of elevation changes to 30˚. If the aeroplane is flying at a constant height of 1500√3 m, find the speed of the plane in km/hr.

#### Solution

Let BC be the height at which the aeroplane is observed from point A.

Then, BC = 1500√3

In 15 seconds, the aeroplane moves from point A to D.

A and D are the points where the angles of elevation 60 and 30

are formed respectively.

Let BA = x metres and AD y metres

BC = x + y

In ΔCBA,

`tan60^@="BC"/"BA"`

`sqrt3=(1500sqrt3)/x`

x=1500 m........(1)

In ΔCBD,

`tan30^@="BC"/"BD"`

`1/sqrt3=(1500sqrt3)/(x+y)`

x + y =1500(3)= 4500

1500 + y = 4500

y = 3000 m ....(2)

We know that the aeroplane moves from point A to D in 15 seconds and the distance covered is 3000 metres. (by 2)

`"speed"="distance"/"time"`

`"speed"=3000/15`

speed=200 m/s

Converting it to km/hr =200x(18/5)=720 km/hr