Answer
$49^{\circ}$
Work Step by Step
We know that
$I_4=\frac{1}{2}I_{\circ}cos^6\theta$
But given that $I_4=\frac{1}{25}I_{\circ}$
$\implies \frac{1}{25}I_{\circ}=\frac{1}{2}I_{\circ}cos^6\theta$
This simplifies to:
$cos\theta=(\frac{2}{25})^{\frac{1}{6}}$
$\implies \theta=cos^{-1}(\frac{2}{25})^{\frac{1}{6}}$
$\theta=49^{\circ}$
