Answer :

Let AE be the tower and CD be the observer. Join E and D. The observer is 28.5 m away from the tower. So, ED = 28.5 m. The observer is 1.5 m tall. So, CD = 1.5 m. Given that the angle of elevation of the top of the tower from the eye of the observer is 45°. Join A and C. Also draw a line BC from C onto AE parallel to DE. We get a right-angled triangle ABC with right angle at B and ∠ ACB = 45°. We are to find the height of the tower, that is, AE.

We see that, BC = ED = 28.5 m.

In ∆ABC,

or,

So, the height of the tower is AE = EB + BA = 1.5m + 28.5m = 30m.

The correct option is (B).

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