Answer
$15^{\circ}$
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, $\theta=\tan^{-1} \dfrac{1-\dfrac{1}{\sqrt 3}}{1+(1)(\dfrac{1}{\sqrt 3})}$
or, $\theta=15^{\circ}$
