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^circ`

⇒ P₁B = AB

In right-angled ΔP₂AB, 

⇒ `(P_2B)/(AB) = tan 60^circ = sqrt(3)`

⇒ `AB = (P_2B)/sqrt(3)`

⇒ `AB = 3000/sqrt(3)`

⇒ `AB = 1000sqrt(3)`

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

= `3000 - 1000sqrt(3)`

= `1000( 3 - sqrt(3))` m