The angle of elevation of an aeroplane from a point on the ground is 60°. After a flight of 30 seconds the angle of elevation becomes 300 If the aeroplane is flying at a constant height of 3000 3 m, find the speed of the aeroplane.

#### Solution

Let P and Q be the two positions of the plane and A be the point of observation. Let ABC be the horizontal line through A.

It is given that angles of elevation of the plane in two positions P and Q from a point A are 60° and 30° respectively.

∠PAB = 60°, ∠QAB = 30°. It is also given that PB = 3000√3 meters

In ΔABP, we have

`tan 60="BP"/"AB"`

`sqrt3/1=(3000sqrt3)/"AB"`

AB=3000m

In ΔACQ, we have

`tan 30="QC"/"AC"`

`1/sqrt3=(3000sqrt3)/"AC"`

AC = 9000 m

∴ Distance = BC = AC – AB = 9000m – 3000m = 6000m

Thus, the plane travels 6km in 30 seconds

Hence speed of plane = 6000/30 = 200 m/sec = 720km/h