Answer
$\theta=cos^{-1}\frac{1}{\sqrt 5}=63.4^{o}$
Work Step by Step
We can find the required angle as follows:
$I_1=\frac{1}{2}I_{\circ}$
and $I_2=\frac{1}{2}I_{\circ}cos^2\theta=\frac{1}{10}I_{\circ}$
Now $\theta =cos^{-1}\sqrt{\frac{2}{I_{\circ}}(\frac{1}{10}I_{\circ})}$
$\implies \theta=cos^{-1}\frac{1}{\sqrt 5}=63.4^{o}$
