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 30°. If the aeroplane is flying at a constant height of 3000√3m, find the speed of the aeroplane.

Let E and D 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 E and D are a point A are 60° and 30° respectively.
∠EAB = 60°, ∠DAB = 30°.
It is also given that EB = 3000√3m.
In ΔABE, tan60° = BE/AB
√3/1 = 3000√3/AB
AB = 3000m
In ΔACD, tan30° = DC/AC
1/√3 = 3000√3/AC
AC = 9000m
Distance = BC = AC – AB = 9000 – 3000 = 6000m
The plane travels 6km in 30 seconds,
Then the speed of the plane = 6000/30 = 200m/sec