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Answer :
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.