Let AB be the tower and C be the point on the ground 20 m away from the foot of the tower B from where the angle of elevation of the top of the tower is 60°. Now, draw a line from C to A. Join B and C. We get a triangle ABC with right angle at B. We are to find the height of the tower, that is AB. We will be using trigonometric ratios involving AB(height) and BC(base). So, ∠ACB = 60° BC = 20m.
Now in ∆ABC,
AB = 20√3 = 34.64m.
The height of the tower is 34.64 m.
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