Answer
$\theta=15^{\circ}$
Work Step by Step
We know that $\tan \theta=\dfrac{m_2-m_1}{1+m_1m_2}$
or, $\theta=tan^{-1} \dfrac{m_2-m_1}{1+m_1m_2}$
We need to plug data into the above expression.
Now, $\theta=\tan^{-1} \dfrac{\sqrt 3-1}{1+(1)(\sqrt 3)}$
or, $\theta=15^{\circ}$
