From a point P on a level ground, the angle of elevation of the top tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is:

Question

Accepted Answer

Let AB be the tower. Then, ∠APB = 30° and AB = 100 m. AB = tan 30° = 1 AP √3

⟹ AP	= (AB x √3) m
	= 100√3 m
	= (100 x 1.73) m
	= 173 m.

From a point P on a level ground, the angle of elevation of the top tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is:

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