question_answer

The angles of depression of two ships from the top of the light house are \[{{45}^{o}}\] and \[{{30}^{o}}\] towards east. If the ships are \[100\,\,m\] apart. then the height of the light house is:

A) \[50(\sqrt{3}-1)m\]                     

B) \[50(\sqrt{3}+1)m\]  

C) \[50(\sqrt{3})m\]

D)        \[\frac{50}{\sqrt{3}-1}m\]

Correct Answer: B

Solution :

 Here let \[AB\] be the height of light house and D, C are the ships such that\[CD=100\,\,m\]. Applying short cut method.             \[AB=\frac{100}{\cot {{30}^{o}}-\cot {{45}^{o}}}\] \[\Rightarrow \]   \[AB=\frac{100}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}\]             \[AB=\frac{100(\sqrt{3}+1)}{2}\]             \[=50(\sqrt{3}+1)m\]