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Q16. The angle of elevation of the top of a tower from two points at a distance of 4 m and 9m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Answer :


Let AB be the tower. Let C and D be the two points at distances 9 m and 4 m respectively from the base of the tower. Then AC = 9 m, AD = 4m.



Let \( ∠ \ ACB \ = \ \theta \) and \( ∠ \ ADB \ = \ 90º \ - \ \theta \)

Let h be the height of the tower AB.



In \( ∆ \ ACB \), we have

\( \frac{AB}{AC} \ = \ tan \theta \ => \ \frac{h}{9} \ = \ tan \theta \)      (1)

In \( ∆ \ ADB \), we have

\( \frac{AB}{AD} \ = \ tan(90º \ - \ \theta) \ => \ \frac{h}{4} \ = \ cot \theta \)      (2)

From (1) and (2), we have

\( \frac{h}{9} \ × \ \frac{h}{4} \ = \ tan \theta \ × \ cot \theta \ => \ \frac{h^2}{36} \ = \ 1 \) \( => \ h^2 \ = \ 36 \ => \ h \ = \ 6 \)

∴ the height of the tower is 6 metres.