question_answer

The angle of elevation of a cloud from a point h metres above a lake is \[\theta \]. The angle of depression of its reflection in the lake is\[{{45}^{o}}\]. The height of the cloud is ____.

A)  \[\frac{h(1+\tan \theta )}{(1-\tan \theta )}\]

B)  \[\frac{h(1-\tan \theta )}{1+\tan \theta }\]

C)  \[h\,\tan \,({{45}^{o}}-\theta )\]

D)    None of these

Correct Answer: A

Solution :

Let A be position of cloud and height of cloud be H metres. In  \[\Delta \,ABC,\,\cot \theta =\frac{BC}{H-h}\] \[(H-h)\,cot\theta =BC\]                      ?..(i) Also, in \[\Delta BCD,\,\,\cot {{45}^{p}}=\frac{BC}{H+h}\] \[\Rightarrow \]   \[BC=(H+h)\,\,\cot {{45}^{o}}\]     ...(ii) From (i) and (ii), we get \[(H-h)\,\cot \theta =(H+h)cot{{45}^{o}}\] \[\Rightarrow \]   \[\frac{H+h}{H-h}=\frac{\tan {{45}^{o}}}{\tan \theta }\] Applying componendo and dividendo, we get (ii) \[\Rightarrow \] \[\frac{H+h+H-h}{H+h-H+h}=\frac{\tan {{45}^{o}}+\tan \theta }{\tan {{45}^{o}}-\tan \theta }\] \[\Rightarrow \]    \[H=h\left( \frac{1+\tan \theta }{1-\tan \theta } \right)\]