Question

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

• A. $\frac{3-2\sqrt{2}}{2}$
• B. $\frac{3+\sqrt{2}}{2}$
• C. $\frac{3-2\sqrt{2}}{4}$
• D. $\frac{3-\sqrt{2}}{4}$

Answer 1

Answer: (C)

Let $BC=CQ=x$ & $AB=h$ and $PQ=2h$
$\tan \theta =\frac{h}{x},\tan \frac{\pi }{8}=\frac{2h}{x}$
$\frac{\tan \theta }{\tan \left(\frac{\pi }{8}\right)}=\frac{1}{2}$
$\tan \theta =\frac{1}{2}\tan \left(\frac{\pi }{8}\right)=\frac{1}{2}(\sqrt{2}-1)$
${\tan }^2\theta =\frac{1}{4}(3-2\sqrt{2})$