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Question
From the top of a 50 m high tower, the angles of depression of the top and bottom of a pole are observed to be 45° and 60° respectively. Find the height of the pole.
Solution
Let H be the height of the pole, makes an angle of depression from the top of the tower to top and bottom of\ poles are 45° and 60° respectively.
Let AB = H , CE = h, AD = x and DE = 50m.
`∠CBE = 45^@ and ∠DAE = 60^@`
Here we have to find height of pole.
The corresponding figure is as follows
In ΔADE
`=> tan A = (DE)/(AD)`
`=> tan 60^@ = 50/x`
`=> x = 50/sqrt3`
Again in ΔBCE
`=> tan B = (CE)/(BC)`
`=> tan 45^@ = h/x`
`=> 1 = h/x`
`=> h = 50/sqrt3`
`=> h = 28.87`
Therefore H = 50 - h
=> H = 50 - 28.87
`=> H = 21.13`
Hence height pole is 21.13 m
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