An observer 1.6 m tall is 20√3 away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The heights of the tower is:

21.6 m
23.2 m
24.72 m
None of these

Correct Answer: 21.6 m

Explanation:

Let AB be the observer and CD be the tower.

Draw BE ⊥ CD.

Then, CE = AB = 1.6 m,

BE = AC = 20√3 m.

DE	= tan 30° =	1
BE	= tan 30° =	√3

⟹ DE =	20√3	m = 20 m.
	√3

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

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