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Question
An observer, 1.5 m tall, is 28.5 m away from a tower 30 m high. Determine the angle of elevation of the top of the tower from his eye.
Solution
Let BE be the observer of 1.5 m tall. And AD be the tower of height 30. Here we have to find the angle of elevation of the top of the tower.
Let ∠ABC = θ
The corresponding figure is as follows
In ΔABC
`=> tan theta = (AC)/(BC)`
`=> tan theta = 28.5/28.5`
`=> tan theta = 1`
`=> theta = 45^@`
Hence the required angle is 45°
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