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Question
An observer finds the angle of elevation of the top of the tower from a certain point on the ground as 30°. If the observe moves 20 m towards the base of the tower, the angle of elevation of the top increases by 15°, find the height of the tower.
Solution
Let the observer be at point C on the ground.
∴ ∠C = 30°
He moves 20 m towards the tower and reaches point D.
∴ ∠D = 30° + 15° = 45°
Let the distance BD be x m and the height of the tower be h m.
In ΔABD
`tan 45^@ = "AB"/"BD" = h/x`
`=> 1 = h/x`
`=> h = x`
In ΔABC
`tan 30^@ = (AB)/(BC) = h/(x + 20) = h/(h + 20)`
`=> 1/sqrt3 = h/(h + 20)`
`=> h + 20 = sqrt3h`
`=> h = 20/(sqrt3 - 1) =10(sqrt3 + 1) m`
Hence, the height of the tower is `10(sqrt3+1)` m
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