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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.