Answer
$\theta = 30^{\circ}$
Work Step by Step
Let $a = \sqrt{3}$, let $b=1$, and let $c = 1$.
We can use the law of cosines to find $\theta$:
$c^2 = a^2+b^2-2ab~cos~\theta$
$2ab~cos~\theta = a^2+b^2-c^2$
$cos~\theta = \frac{a^2+b^2-c^2}{2ab}$
$\theta = arccos(\frac{a^2+b^2-c^2}{2ab})$
$\theta = arccos(\frac{(\sqrt{3})^2+1^2-1^2}{(2)(\sqrt{3})(1)})$
$\theta = arccos(\frac{3}{2\sqrt{3}})$
$\theta = arccos(\frac{\sqrt{3}}{2})$
$\theta = 30^{\circ}$
