question_answer

Form the top of a light house \[60\,\,m\] high with its base at the sea level, the angle of depression of a boat is\[{{15}^{o}}\]. The distance of the boat form the foot of the light house is

A) \[\left( \frac{\sqrt{3}-1}{\sqrt{3}+1} \right)60\,\,m\]      

B) \[\left( \frac{\sqrt{3}+1}{\sqrt{3}-1} \right)60\,\,m\]

C) \[\left( \frac{\sqrt{3}+1}{\sqrt{3}-1} \right)m\]                 

D)  None of these

Correct Answer: B

Solution :

Let \[OA\] be the light house and \[B\] be the position of boat. It is given that\[LOBA=LABL={{15}^{o}}\] Now,\[\tan {{15}^{o}}=\frac{OA}{OB}\] \[\Rightarrow \]               \[OB=OA\cdot \cot {{15}^{o}}\] \[\Rightarrow \]               \[OB=60\cdot \cot ({{45}^{o}}-{{30}^{o}})\] \[=60\left( \frac{\cot {{45}^{o}}\cdot \cot {{30}^{o}}+1}{\cot {{30}^{o}}-\cot {{45}^{o}}} \right)=60\left( \frac{\sqrt{3}+1}{\sqrt{3}-1} \right)m\].