## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$150^{\circ}$
Let us consider that a vector $v$ is given by $v=pi+qj$ and makes an angle of $\alpha$ with the positive $x$-axis. The direction angle $\alpha$ for the given vector can be determined using the formula: $\tan \alpha=\dfrac{q}{p}$ Here, we have: $v=-3 \sqrt 3i +3 j$ Now, $\tan \alpha=\dfrac{ 3}{-3 \sqrt 3}=-\dfrac{\sqrt 3}{3}$ Since, $0\leq \alpha \leq 360^{\circ}$ Therefore, the direction angle is: $\alpha=\arctan (-\dfrac{\sqrt 3}{3})=150^{\circ}$