Question

An aeroplane when flying at a height of 3000 m from the ground  passes vertically above another aeroplane at an instant when the angles of elevation of the two planes from the same point on the ground are 60° and 45° respectively. Find first aeroplance is how much high with second aeroplane. 

Answer 1

Let height of first aeroplane is BD which is flying at 3000 m height and height of second aeroplane BC = h m. 

Let angles of elevation from point A are 60° and 45° respectively. 

So,
∠BAD = 60° and ∠CAB = 45°and 

Let AB = x m.

From right angled ΔABD,

tan 60° = BD/AB

⇒ √3 = 3000/x

Hence, height of second plane from first plane

CD = BD – BC 

= 3000 – 1732 = 1268 m