Question 141

From a point 20 m away from the foot of a tower, the angle of elevation of the top of the tower is 30°. The height of the tower is

Solution

Perpendicular AB= height of tower

Base BC = distance b/w base of tower and the point

Hypotenuse AC= segment b/w top of the tower and the point

Angle of elevation ACB =30°

tan 30° = Perpendicular / Base = 1/√3

Perpendicular = Base*1/√3

Base is 20 m.

So height of tower is 20/√3m.

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