An observer 1.6 m tall is 203 away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The height of the tower is:

Answer: Option A

Explanation:

Let AB be the observer and CD be the tower.

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Draw BE CD.

Then, CE = AB = 1.6 m,

BE = AC = 203 m.

DE	= tan 30° =	1
BE	= tan 30° =	3

DE =	203	m = 20 m.
	3

CD = CE + DE = (1.6 + 20) m = 21.6 m.