A tower stands vertically on the ground. From a point on the ground which is 20 m away from the foot of the tower, the angle of elevation of its top is found to be 60°. Find the height of the tower. [Take √3 = 1.732.]

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,

or,

AB = 20√3 = 34.64m.

The height of the tower is 34.64 m.