question_answer

Two poles are 10 m and 20m high. The line joining their tops makes an angle of \[15{}^\circ \] with the horizontal. The distance between the poles is approximately equal to

A) \[36.3m\]

B) \[37.3in\]

C) \[38.3m\]

D) \[39.3in\]

Correct Answer: B

Solution :

[b] \[\tan 15{}^\circ =DE/AE\]

\[AE=10\,\,cot\,\,15{}^\circ \]                ? (1)

\[\cot (15{}^\circ )=cot(45-30{}^\circ )\]

\[=\frac{\cot 45{}^\circ \cot 30{}^\circ +1}{\cot 30{}^\circ -\cot 45{}^\circ }\]

\[\cot 15{}^\circ =\frac{1.\sqrt{3}+1}{\sqrt{3}-1}=2+\sqrt{3}\]

Putting \[\cot 15{}^\circ \] in eq (1)

\[AE=10\cot 15{}^\circ \]

\[=10(2+\sqrt{3})\]

\[=10(3.73)\]

\[=37.3m\]