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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?
Solution
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
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