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The polar co-ordinates of the point, whose Cartesian co-ordinates are $(-2 \sqrt{3}, 2)$, are
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$\left(4,\left(\frac{5 \pi}{6}\right)\right)$
$\begin{aligned} & x=-2 \sqrt{3} \text { and } y=2 \\ & r=\sqrt{x^2+y^2}=\sqrt{12+4}=4 \\ & \theta=\pi-\tan ^{-1}\left|\frac{2}{-2 \sqrt{3}}\right|=\pi-\tan ^{-1} \frac{1}{\sqrt{3}}=\pi-\frac{\pi}{6}=\frac{5 \pi}{6}\end{aligned}$
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