Let $AB$ and $PQ$ be two vertical poles, 160 m 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\\
(1) $\frac{3-2\sqrt{2}}{2}$\\
(2) $\frac{3+\sqrt{2}}{2}$\\
(3) $\frac{3-2\sqrt{2}}{4}$\\
(4) $\frac{3-\sqrt{2}}{4}$