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:

Question

Accepted Answer

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…

DE	= tan 30° =	1
BE	= tan 30° =	√3

⟹ DE =	20√3	m = 20 m.
	√3

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