What are the polar coordinates of the point ( | vedclass

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(A) Given point is $(x, y) = (-\sqrt{3}, 1)$.To find polar coordinates $(r, 	heta)$,we use $r = \sqrt{x^2 + y^2}$.$r = \sqrt{(-\sqrt{3})^2 + (1)^2} =

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$(2, 5\pi/6)$

Vedclass · Answer

(A) Given point is $(x, y) = (-\sqrt{3}, 1)$.To find polar coordinates $(r, 	heta)$,we use $r = \sqrt{x^2 + y^2}$.$r = \sqrt{(-\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.Since the point $(-\sqrt{3}, 1)$ lies in the second quadrant,the angle $	heta$ is given by $	an 	heta = y/x = 1/(-\sqrt{3})$.Thus,$	heta = \pi - \pi/6 = 5\pi/6$.Therefore,the polar coordinates are $(2, 5\pi/6)$.

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

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