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.
In the fig let C be the initial position of the aeroplane. After 15 seconds the position of the aeroplane becomes E.
In ∆ABC
tan 45° =
1 =
Y = 3000m ------(2)
In ∆ADE
tan 30° =
=
x + y = 3000
Using equation (1) to replace value of y, we get
x = 3000√3 – 3000
⇒ 3000(√3-1)
⇒ 2196m
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