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√3 m, find the speed of the aeroplane.

Question

Philoid · Accepted Answer

Let P and Q be the two positions of the aeroplane 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° and ∠QAB = 30°

Also given PB = 3000√3 m

In Δ ABP,

⇒ tan 60° = BP/AP

⇒ √3

⇒ AB = 3000 m

In Δ ACQ,

⇒ tan 30° = CQ/AC

⇒

⇒ AC = 3000 (3)

⇒ AC = 9000 m

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

Thus, the plane travels 6000m in 30 seconds.

We know that speed = distance/time

⇒ Speed of plane = 200 m/sec

∴ Speed of the aeroplane is 200 m/sec.