How do you convert $(-2, 2sqrt3)$ to polar form?

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Answer 1 · 2016-05-22T11:16:16

In Polar form $(r,theta)$ is $4,120^0$

Polar distance $r=sqrt((-2)^2+(2sqrt3)^2)=4$
$theta=tan^-1((2sqrt3)/-2)=tan^-1(- sqrt3) =120^0$ since the point is on 2nd quadrant [Ans]

How do you convert (-2, 2sqrt3) (-2, 2sqrt3) to polar form?