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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Eric Bizzell
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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...
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