Explanation

(c)
Given, $AD = p$ and $BC = 2\sqrt{2}\ p$
Clearly, $p = {a\ sin}{\theta =}b\cos\theta$
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Since, $a^{2} + b^{2} = \left( 2\sqrt{2}p \right)^{2}$
$\Rightarrow p^{2}\left\lbrack \frac{1}{\sin^{2}\theta} + \frac{1}{\cos^{2}\theta} \right\rbrack = 8p^{2}$
$\Rightarrow \ \ 1 = 2\sin^{2}{2\theta}\ \ \ \Rightarrow \ \sin{2\theta = \pm \frac{1}{\sqrt{2}}}$
$\Rightarrow \sin{2\theta} = \frac{1}{\sqrt{2}}\ \ \Rightarrow \ \ 2\theta = \frac{\pi}{4}\ \Rightarrow \ \theta = \frac{\pi}{8}$
So, the other angle is $\frac{\pi}{2} - \theta = \frac{\pi}{2} - \frac{\pi}{8} = \frac{3\pi}{8}$