## Precalculus (10th Edition)

$(-3\sqrt3,3)$
By definition: $x=r\cdot \cos(\theta)$ and $y=r\cdot \sin(\theta)$. Here $r=6 \text{ and } \theta=150^\circ$. Hence, after plugging these in: $x=6\cdot \cos(150^\circ)=6\cdot(-\frac{\sqrt3}{2})=-3\sqrt3$ $y=6\cdot \sin(150^\circ)=6\cdot(\frac{1}{2})=3$ The point has coordinates $(-3\sqrt3,3)$ in the rectangular coordinate system.