## Precalculus (6th Edition)

$30^0$
From the law of cosines $1^{2}=1^{2}+\left( \sqrt {3}\right) ^{2}-2\times 1\times \left( \sqrt {3}\right) \times \cos \theta \Rightarrow 2\sqrt {3}\cos \theta =3\Rightarrow \cos \theta =\dfrac {\sqrt {3}}{2}\Rightarrow \theta =\cos ^{-1}\left( \dfrac {\sqrt {3}}{2}\right) =30^0$