Answer
The polar coordinates of $\left( -2\sqrt{3},2 \right)$ are $\left( 4,\frac{5\pi }{3} \right)$.
Work Step by Step
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 $\left( 4,\frac{5\pi }{3} \right)$.
