What are the polar coordinates of (-sqrt{3}, | vedclass

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Question

(C) Given the Cartesian coordinates $(x, y) = (-\sqrt{3}, -1)$.We know that $r = \sqrt{x^2 + y^2} = \sqrt{(-\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = \sq

Vedclass · Accepted Answer

$(2, -5\pi/6)$

Vedclass · Answer

(C) Given the Cartesian coordinates $(x, y) = (-\sqrt{3}, -1)$.We know that $r = \sqrt{x^2 + y^2} = \sqrt{(-\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.For the angle $	heta$,we have $\cos 	heta = \frac{x}{r} = \frac{-\sqrt{3}}{2}$ and $\sin 	heta = \frac{y}{r} = \frac{-1}{2}$.Since both $\sin 	heta$ and $\cos 	heta$ are negative,the point lies in the third quadrant.The reference angle is $\alpha = 	an^{-1}\left|\frac{-1}{-\sqrt{3}}ight| = 	an^{-1}\left(\frac{1}{\sqrt{3}}ight) = \frac{\pi}{6}$.In the third quadrant,$	heta = -(\pi - \frac{\pi}{6}) = -\frac{5\pi}{6}$.Thus,the polar coordinates are $(2, -5\pi/6)$.

What are the polar coordinates of $(-\sqrt{3}, -1)$?

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