Answer
a) $L=\sqrt{l_{\circ}^2(1-\frac{v^2}{c^2}cos^2\theta_{\circ})}$
b) $\theta=tan^{-1}(\frac{tan\theta}{\sqrt{1-\frac{v^2}{c^2}}})$
Work Step by Step
(a) We know that
$L=\sqrt{L_x^2+L_y^2}$
$\implies L=\sqrt{(l_{\circ}cos\theta_{\circ}\sqrt{1-\frac{v^2}{c^2}})^2+(l_{\circ}sin\theta_{\circ})^2}$
$\implies L=\sqrt{(l_{\circ}cos\theta_{\circ})^2(1-\frac{v^2}{c^2})+(l_{\circ}sin\theta_{\circ})^2}$
This simplifies to:
$L=\sqrt{l_{\circ}^2(1-\frac{v^2}{c^2}cos^2\theta_{\circ})}$
(b) We know that
$\theta=tan^{-1}(\frac{L_y}{L_x})$
$\theta=tan^{-1}(\frac{l_{\circ}sin\theta_{\circ}}{(l_{\circ}cos\theta_{\circ})\sqrt{1-\frac{v^2}{c^2}}})$
This simplifies to:
$\theta=tan^{-1}(\frac{tan\theta}{\sqrt{1-\frac{v^2}{c^2}}})$
