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Question
An aeroplane when flying at a height of 4km from the ground passes vertically above another aeroplane at an instant when the angles of the elevation of the two planes from the same point on the ground are 60° and 45° respectively. Find the vertical distance between the aeroplanes at that instant.
Solution
Let points A and D represent the position of the aeroplanes.
Aeroplane A is flying 4 km = 4000 m above the ground.
∠ACB = 60°, ∠DCB = 45°
In ΔABC,
`"AB"/"BC" = tan 60^circ`
⇒ `"BC" = 4000/sqrt(3)`
In ΔDCB,
`"DB"/"BC" = tan 45^circ`
⇒ DB = BC = `4000/sqrt(3)`
∴ AD = AB - BD
= `4000 - 4000/sqrt(3) = 4000(1 - 1/sqrt(3)) = 4000 xx (sqrt(3) - 1)/sqrt(3) = 4000 xx 0.732/1.732 = 1690.53`
= h = `sqrt(3)"x" = 1.732 xx 136.6 = 236.59 ≈ 236.6` m
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