How do you convert (-5, 6) into polar coordinates?

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Answer 1 · 2015-11-27T02:19:28

Find #r and theta# using the following ...

#r=sqrt(-5^2+6^2)=sqrt61#

Now, the coordinate is in Quadrant II , so
#90^o < theta < 180^o#

#theta=arctan(6/5)=50.19^o#, this is called the reference angle

Since we know it lies in Quadrant II ...

#theta = 180-50.19=129.81^o#

Polar Coordinate #=(sqrt61, 129.81^o)# or #(sqrt61, 0.72pi)#

hope that helped