Let AB and PQ be two vertical poles, 160m apart from each other....

From the given information in the question we can draw the following diagram

Let $BC = CQ = x \text{ and } AB = h$ and $PQ = 2h$

So $tan\theta = \dfrac{h}{x}$ and $tan\dfrac{\pi}{8} = \dfrac{2h}{x}$

$⇒ \dfrac{tan\theta}{\tan \left(\dfrac{\pi}{8}\right)} = \dfrac{1}{2}$

$⇒ tan\theta = \dfrac{1}{2}\left(\sqrt{2} - 1\right)$

$⇒ \tan^{2}\theta = \dfrac{1}{4}\left(3 - 2\sqrt{2}\right)$