Question

An aeroplane when 3,000 meters high passes vertically above another aeroplane at an instance when their angles of elevation at the same observation point are 60° and 45° respectively. How many meters higher is the one than the other?

Answer 1

Let P₁ and P₂ denote the positions of the two planes. Then in right-angled ΔP₁AB, 

`(P_1B)/(AB) = tan 45° ⇒ P_1B = AB`  

In right-angled ΔP₂AB, 

⇒ `(P_2B)/AB = tan 60° = sqrt3`

⇒ AB = `(P_2B)/sqrt3`

⇒ AB = `3000/sqrt3`

⇒ AB = 1000√3

∴ Vertical distance between the two planes is P₁P₂ = P₂B - P₁B

= 3000 - 1000√3
= 1000( 3 - √3) m