[tex]\bf \begin{array}{clclll}
(-3&,&\sqrt{3})\\
x&&y
\end{array}\qquad
\begin{cases}
r=\sqrt{x^2+y^2}\\\\
\theta=tan^{-1}\left( \frac{y}{x} \right)
\end{cases}\\\\
-----------------------------\\\\[/tex]
[tex]\bf r=\sqrt{(-3)^2+(\sqrt{3})^2}\implies r=\sqrt{9+3}\implies r=\sqrt{12}\implies r=2\sqrt{3}
\\\\\\
\theta=tan^{-1}\left( \frac{\sqrt{3}}{-3} \right)
\impliedby
\textit{now, let's rationalize the numerator}
\\\\\\
\cfrac{\sqrt{3}}{-3}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies \cfrac{3}{-3\sqrt{3}}\implies -\cfrac{1}{\sqrt{3}}[/tex]
now, let's take a peek at our terminal point, -3, √(3).... the "x" is negative, the "y" is positive, and that means, the 2nd quadrant
[tex]\bf \textit{now, take a look in your Unit Circle at }\frac{5\pi }{6}\quad
\begin{cases}
x=-\frac{\sqrt{3}}{2}\\\\
y=\frac{1}{2}
\end{cases}
\\\\\\
\cfrac{y}{x}\implies \cfrac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}\implies \cfrac{1}{2}\cdot -\cfrac{2}{\sqrt{3}}\implies -\cfrac{1}{\sqrt{3}}
\\\\\\
\theta=\cfrac{5\pi }{6}
\\\\\\
thus\qquad \qquad \left( 2\sqrt{3}\ ,\ \frac{5\pi }{6} \right)[/tex]
