## Precalculus (10th Edition)

$(3, 0)$
We know that if the rectangular coordinates of the point are $(x,y)$, then $r=\sqrt{x^2+x^2}$ and $\theta=\tan^{-1}(\frac{y}{x})$. Hence here: $r=\sqrt{3^2+0^2}=3$ $\theta=\tan^{-1}\left(\frac{0}{3}\right)=0$ The given point has the polar coordinates $(3,0)$.