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 \[={{\theta }_{1}}+{{\theta }_{2}}+{{\theta }_{3}}\] \[=\frac{\pi }{4}+{{\tan }^{-1}}\left( \frac{1}{2} \right)+{{\tan }^{-1}}\left( \frac{1}{3} \right)\] \[=\frac{\pi }{4}+{{\tan }^{-1}}\left( \frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{3}\times \frac{1}{2}} \right)\] \[=\frac{\pi }{4}+{{\tan }^{-1}}\left( \frac{5/6}{5/6} \right)\] \[=\frac{\pi }{4}+{{\tan }^{-1}}(1)=\frac{\pi }{4}+\frac{\pi }{4}=\frac{\pi }{2}\]You need to login to perform this action.
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