# From the top of a cliff 20 m high, the angle of elevation of the top of a tower is found to be equal to the angle of depression of the foot of the tower. The height of the tower isA. 20 mB. 40 mC. 60 mD. 80 m

Let BD be the cliff and AE be the tower. Join D and E. Now, given that the angle of elevation of the top of the tower is same as the angle of depression of the foot of the tower as seen from the top of the cliff. Join A with B and E with B. Also draw a line BC from B onto AE parallel to DE. We get two right-angled triangles ABC and BDE with right angles at C and D respectively. The angle of elevation of the top of the tower AE from the point B is ABC. And the angle of depression of the foot of the tower AE from the point B is CBE which is same as BED. Given, BD = 20m. We are to find the height of the tower, that is BD. Clearly, BD = CE = 20m, BC = DE. In ∆BCE,

or,

In ∆ABC,

or,

AC = 20m

So, height of the tower is AE = AC + CE = 20m + 20m = 40m

The correct option is (B).

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