# What are the polar coordinates (r, theta) of the point (-1,-square root of 3) where r > 0 and 0 < (or equal to) theta < 2pi.

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### 1 Answer

Given the point `(-1,-sqrt(3))` in rectangular coordinates, we are asked to convert to polar coordinates with `r>0,0<=theta<2pi` :

The conversions are `r=sqrt(x^2+y^2)` and `theta=tan^(-1)(y/x)` where you must take into account the quadrant.

** If you plot the point and draw a perpendicular segment to the x-axis you will form a right triangle with the hypotenuse the segment from the point to the origin, and the legs the segment drawn perpendicular to the x-axis and the segment along the x-axis.

Thus we use the Pythagorean theorem to find r, and the fact that the tangent of theacute angle with its vertex at the origin is `y/x` **

Here `r=sqrt((-1)^2+(-sqrt(3))^2)=sqrt(4)=2` and

`theta=tan^(-1)(-sqrt(3))/(-1))=tan^(-1)sqrt(3)=pi/3` ; but since the point is in the third quadrant we have `theta=(4pi)/3`

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The polar coordinates are `(2,(4pi)/3)`

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