Trigonometry (11th Edition) Clone

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