question_answer

The length of the shadows of a vertical pole of height\[h\], thrown by the sun's rays at three different moments are \[h,\,\,2h\] and\[3h\]. The sum of the angles of elevation of the rays at these three moments is equal to

A) \[\frac{\pi }{2}\]                                              

B) \[\frac{\pi }{3}\]

C) \[\frac{\pi }{4}\]                                              

D)  \[\frac{\pi }{6}\]

Correct Answer: A

Solution :

Let \[OA\] be the vertical pole of height \[h\] and \[O{{P}_{1}},\,\,O{{P}_{2}}\]\[\text{and}\,\,O{{P}_{3}}\] be the lengths of shadow. In \[\Delta AOP\] we have                 \[\tan {{\theta }_{1}}=\frac{OA}{O{{P}_{1}}}=\frac{h}{h}=1\Rightarrow {{\theta }_{1}}=\frac{\pi }{4}\] In\[\Delta AO{{P}_{2}}\], we have                 \[\tan {{\theta }_{2}}=\frac{OA}{O{{P}_{2}}}=\frac{h}{2h}=\frac{1}{2}\] Similarly, in\[\Delta AO{{P}_{3}}\], we have                 \[\tan {{\theta }_{3}}=1/3\]                 \[{{\theta }_{3}}={{\tan }^{-1}}(1/3)\] Sum of the angles of elevation of the rays                 \[={{\theta }_{1}}+{{\theta }_{2}}+{{\theta }_{3}}\]                 \[=\frac{\pi }{4}+{{\tan }^{-1}}\left( \frac{1}{2} \right)+{{\tan }^{-1}}\left( \frac{1}{3} \right)\]                 \[=\frac{\pi }{4}+{{\tan }^{-1}}\left( \frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{3}\times \frac{1}{2}} \right)\]                 \[=\frac{\pi }{4}+{{\tan }^{-1}}\left( \frac{5/6}{5/6} \right)\]                 \[=\frac{\pi }{4}+{{\tan }^{-1}}(1)=\frac{\pi }{4}+\frac{\pi }{4}=\frac{\pi }{2}\]