Question

The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of 1500 √3 m, find the speed of the jet plane.

Answer 1

Let P and Q be the two positions of the plane and let 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.
Then, ∠PAB = 60°, ∠QAB = 30°
It is also given that PB = 1500 √3 metres
In ΔABP, we have
tan 60° = BP/AB m
⇒ √3 = 1500 √3/AB
⇒ AB = 1500
In ΔACQ, we have
tan 30° = CQ/AC ⇒ 1/√3 = 1500 √3/AC
⇒ AC = 1500 × 3 = 4500 m

PQ = BC = AC – AB = 4500 – 1500 = 3000 m
Thus, the plane travels 3000 m in 15 seconds.
Hence, the speed of plane = 3000/15 = 200 = 200 m/s
= 200 × 3600 /1000 km/h = 200 ×18/5 km/h = 720 km/h.