## Algebra 2 (1st Edition)

$30^{\circ}$ or, $\dfrac{\pi}{6}$
Here, we have $\tan^{-1} (\dfrac{\sqrt 3}{3})$ This means that $\theta=\tan^{-1} (\dfrac{\sqrt 3}{3})$ This gives : $\theta=30^{\circ}$ The range of the function $arccos$ is from $[-90^{\circ}, 90^{\circ}]$. Hence, we have $\theta=30^{\circ}$ or, $\dfrac{\pi}{6}$