Answer
(a) See the graph.
(b) $(6,\frac{\pi}{6})$
(c) $(-6,\frac{7\pi}{6})$
Work Step by Step
(a) $x\gt0$ and $y\gt0$.
The point lies in Quadrant I.
$r^2=x^2+y^2$
$r^2=(3\sqrt 3)^2+3^2=36$
$r=±6$
(b) $r=6$
$tan~θ=\frac{y}{x}=\frac{3}{3\sqrt 3}=\frac{\sqrt 3}{3}$
$θ=\frac{\pi}{6}$
(c) $r=-6$
$tan~θ=\frac{y}{x}=\frac{3}{3\sqrt 3}=\frac{\sqrt 3}{3}$
$θ=\frac{\pi}{6}+\pi=\frac{7\pi}{6}$
