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.

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#### 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

Concept: Heights and Distances

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