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

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Answer 1 · 2015-11-27T02:19:28

Find $r and θ$ $r and theta$ using the following ...

$r = \sqrt{- 5^{2} + 6^{2}} = \sqrt{61}$ $r=sqrt(-5^2+6^2)=sqrt61$

Now, the coordinate is in Quadrant II , so
$90^{o} < θ < 180^{o}$ $90^o < theta < 180^o$

$θ = arctan (\frac{6}{5}) = {50.19}^{o}$ $theta=arctan(6/5)=50.19^o$ , this is called the reference angle

Since we know it lies in Quadrant II ...

$θ = 180 - 50.19 = {129.81}^{o}$ $theta = 180-50.19=129.81^o$

Polar Coordinate $= (\sqrt{61}, {129.81}^{o})$ $=(sqrt61, 129.81^o)$ or $(\sqrt{61}, 0.72 π)$ $(sqrt61, 0.72pi)$

hope that helped