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Let \mathrm{AB} and \mathrm{PQ} be two vertical poles,160 \mathrm{~m} apart from each other. Let \mathrm{C} be the middle point of  \mathrm{B} and \mathrm{Q}, which are feet of these two poles. Let \mathrm{\frac{\pi}{8}} and \mathrm{\theta} be the angles of elevation from \mathrm{C} to \mathrm{P} and \mathrm{A}, respectively. If the height of pole \mathrm{P Q} is twice the height of pole \mathrm{AB}, then \mathrm{\tan ^{2} \theta} is equal to

Option: 1

\frac{3-2 \sqrt{2}}{2}


Option: 2

\frac{3+\sqrt{2}}{2}


Option: 3

\frac{3-2 \sqrt{2}}{4}


Option: 4

\frac{3-\sqrt{2}}{4}


Answers (1)

best_answer

\mathrm{\tan \theta= \frac{h}{x}}  ,   \mathrm{\tan \frac{\pi}{8}= \frac{2h}{x}}

\mathrm{\Rightarrow\frac{\tan \theta}{\tan \frac{\pi}{8}}= \frac{1}{2} }
\mathrm{\Rightarrow \tan \theta= \frac{1}{4}\tan \frac{\pi}{8} }
\mathrm{\Rightarrow \tan ^{2}\theta =\frac{1}{4}\tan ^{2}\frac{\pi}{8}= \frac{1}{4}\left ( \frac{1-\cos \frac{\pi}{4}}{1+\cos \frac{\pi}{4}} \right ) = \frac{1}{4}\left ( \frac{1-1/\sqrt{2}}{1+1/\sqrt{2}} \right )}  \mathrm{= \frac{1}{4}\left ( \frac{\sqrt{2}-1}{\sqrt{2}+1} \right )= \frac{1}{4}\frac{\left ( \sqrt{2}-1 \right )^{2}}{\left ( 2-1 \right )}= \frac{3-2\sqrt{2}}{4} }

Option (C)

Posted by

Pankaj Sanodiya

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