An aeroplane flying horizontally at height of 2500 √ 3 mtsabove that ground; is observed to be at angle of elevation 60° from theground. After a flight of 25 seconds the angle of elevation is 30° . Findthe speed of the plane in m/sec.

Given:

Angle of elevation, θ = 30°

Height above the ground = 2500√ 3 m

To find: speed of the plane

Solution:

Let P and Q be the two positions of the aeroplane.

QB = 2500 √ 3 m

Let the speed of the aeroplane be s m/sec.

tan 60° = 2500√ 3 / x

(2500√ 3 / x) = √ 3

∴ x = 2500 m

tan 30° = ( 2500√ 3) / ( x + y)

( 2500√ 3) / ( x + y) = 1 √ 3

⇒ x + y = 2500 √ 3 × √ 3

= 2500 × 3

= 7500

∴ y = 7500 – 2500

= 5000 m

To travel 5000 m the aeroplane takes 25 seconds.

∴ The speed of the plane = 5000 / 25

= 200 m/sec.