## Precalculus (6th Edition) Blitzer

The polar coordinates of $\left( -2\sqrt{3},2 \right)$ are $\left( 4,\frac{5\pi }{3} \right)$.
The polar coordinates of the point are $\left( r,\theta \right)$. Now, rewrite the polar coordinates in terms of rectangular coordinates as below: $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ ……(1) $\tan \theta =\frac{y}{x}$ ……(2) Substituting the values of $x\ \text{ and }\ y$ in (1) and (2), we get \begin{align} & r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\ & =\sqrt{{{\left( -2\sqrt{3} \right)}^{2}}+{{\left( 2 \right)}^{2}}} \\ & =\sqrt{12+4}=\sqrt{16} \\ & r=4 \end{align} And, \begin{align} & \tan \theta =\frac{y}{x} \\ & =\frac{2}{-2\sqrt{3}} \\ & \tan \theta =-\frac{1}{\sqrt{3}} \end{align} Hence, $\tan \theta =-\frac{1}{\sqrt{3}}$ And, $\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$ Also $\theta$ lies in the second quadrant, which means that \begin{align} & \theta =\pi -\frac{\pi }{6} \\ & \theta =\frac{5\pi }{6} \end{align} Therefore, the polar coordinates of $\left( -2\sqrt{3},2 \right)$ are $\left( 4,\frac{5\pi }{3} \right)$.