If the Cartesian coordinates of a point are l | vedclass

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(C) Given Cartesian coordinates $(x, y) = \left(-\frac{5 \sqrt{3}}{2}, \frac{5}{2}\right)$.$r = \sqrt{x^2 + y^2} = \sqrt{\left(-\frac{5 \sqrt{3}}{2}\r

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$\left(5, \frac{5 \pi}{6}\right)$

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(C) Given Cartesian coordinates $(x, y) = \left(-\frac{5 \sqrt{3}}{2}, \frac{5}{2}ight)$.$r = \sqrt{x^2 + y^2} = \sqrt{\left(-\frac{5 \sqrt{3}}{2}ight)^2 + \left(\frac{5}{2}ight)^2} = \sqrt{\frac{75}{4} + \frac{25}{4}} = \sqrt{\frac{100}{4}} = \sqrt{25} = 5$.Since the point lies in the second quadrant $(x  0)$,the polar angle $	heta$ is given by $	heta = \pi - 	an^{-1}\left|\frac{y}{x}ight|$.$	heta = \pi - 	an^{-1}\left|\frac{5/2}{-5\sqrt{3}/2}ight| = \pi - 	an^{-1}\left(\frac{1}{\sqrt{3}}ight) = \pi - \frac{\pi}{6} = \frac{5 \pi}{6}$.Thus,the polar coordinates are $(r, 	heta) = \left(5, \frac{5 \pi}{6}ight)$.

If the Cartesian coordinates of a point are $\left(\frac{-5 \sqrt{3}}{2}, \frac{5}{2}\right)$,then its polar coordinates are

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