The polar coordinates of the point,whose Cart | vedclass

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(B) Given Cartesian coordinates are $(x, y) = (-2 \sqrt{3}, 2)$.First,calculate the modulus $r$:$r = \sqrt{x^2 + y^2} = \sqrt{(-2 \sqrt{3})^2 + (2)^2}

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$(4, \frac{5 \pi}{6})$

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(B) Given Cartesian coordinates are $(x, y) = (-2 \sqrt{3}, 2)$.First,calculate the modulus $r$:$r = \sqrt{x^2 + y^2} = \sqrt{(-2 \sqrt{3})^2 + (2)^2} = \sqrt{12 + 4} = \sqrt{16} = 4$.Since the point lies in the second quadrant $(x  0)$,the argument $	heta$ is given by:$	heta = \pi - 	an^{-1} \left| \frac{y}{x} ight| = \pi - 	an^{-1} \left| \frac{2}{-2 \sqrt{3}} ight| = \pi - 	an^{-1} \left( \frac{1}{\sqrt{3}} ight) = \pi - \frac{\pi}{6} = \frac{5 \pi}{6}$.Thus,the polar coordinates are $(r, 	heta) = (4, \frac{5 \pi}{6})$.

The polar coordinates of the point,whose Cartesian coordinates are $(-2 \sqrt{3}, 2)$,are

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