A T.V. tower stands vertically on a bank of a river. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60º. From a point 20 m away from this point on the same bank, the angle of elevation of the top of the tower is 30º (see figure). Find the height of the tower and the width of the river.

Let AB be the T.V. tower of height h standing on the bank of a river. Let C be the point on the opposite bank of the river such that BC = x metres. Let D be another point away from C such that CD = 20, and the angles of levation of the top of the T.V. tower at C and D are 60º and 30º respectively. i.e., \( ∠ \ ACB \ = \ 60º \) and \( ∠ \ AOB \ = \ 30º \)



In \( ∆ \ ABC \) , we have

\( \frac{AB}{BC} \ = \ tan60º \)

\( => \ \frac{h}{x} \ = \ \sqrt{3} \ \)
\( => \ h \ = \ \sqrt{3}x \) (1)

In \( ∆ \ ABD \), we have

\( \frac{AB}{BD} \ = \ tan30º \ \)
\( => \ \frac{h}{x+20} \ = \ \frac{1}{ \sqrt{3}} \)
\( => \ h \ = \ \frac{x+20}{ \sqrt{3}} \) (2)

Equating values of h from (1) and (2), we get

\( \sqrt{3}x \ = \ \frac{x+20}{ \sqrt{3}} \ \)
\( => \ 3x \ = \ x + 20 \)
\( => \ x = 10 \)

Putting x = 10 in (1), we get

\( h \ = \ \sqrt{3} × 10 \ = \ 17.32 \)

\(\therefore \) the height of the T.V. tower is 17.32 metres and the widith of the river is 10 metres.