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If the Cartesian co-ordinates of a point are $\left(\frac{-5 \sqrt{3}}{2}, \frac{5}{2}\right)$, then its polar co-ordinates are
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$\left(5, \frac{5 \pi}{6}\right)$
$\begin{aligned} & r=\sqrt{x^2+y^2}=\sqrt{\frac{75}{4}+\frac{25}{4}}=5 \\ & \theta=\pi-\tan ^{-1}\left|\frac{y}{x}\right|=\pi-\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\pi-\frac{\pi}{6}=\frac{5 \pi}{6} \\ & \text { Polar co-ordinate }=(r, \theta)=\left(5, \frac{5 \pi}{6}\right)\end{aligned}$
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