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Answer :
Let AB and CD be two poles each of height h metres. Let P be a point on the road such that AP = x metres. Then CP = (80 – x) metres. Its is given that \( ∠ \ APB \ = \ 60º \) and \( ∠ \ CPD \ = \ 30º \).
In \( ∆ \ APB \), we have
\( \frac{AB}{AP} \ = \ tan60º \ \)
\( => \ \frac{h}{x} \ = \ \sqrt{3} \)
\( => \ h \ = \ \sqrt{3}x \) (1)
In \( ∆ \ CPD \), we have
\( \frac{CD}{CP} \ = \ tan30º \ \)
\( => \ \frac{h}{80-x} \ = \ \frac{1}{ \sqrt{3}} \)
\( => \ h \ = \ \frac{80-x}{ \sqrt{3}} \) (2)
Equating the values of h from (1) and (2), we get
\( \sqrt{3}x \ = \ \frac{80-x}{ \sqrt{3}} \ \)
\( =>\ 3x \ = \ 80-x \)
\( => \ x \ = \ 20 \)
Putting x = 20 in (1), we get
\( h \ = \ \sqrt{3} × 20 \ \)
\( = \ (1.732) × 20 \ = \ 34.64 \)
\(\therefore \) The point is at a distance of 20 metres from the first pole and 60 metres from the second pole. And the height of the pole is 34.64 metres.