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Answer 1 · 2016-11-29T02:44:56

$(r, θ) = (13, arctan (- \frac{12}{5})) \approx (13, - {67.380}^{\circ})$ $(r, theta) = (13, arctan(-12/5)) ~~ (13, -67.380^@)$

There are two main sets of equations when converting between rectangular and polar coordinates. To go from polar to rectangular, we have

${(x = rcos(theta)), (y = rsin(theta)):}$

and to go from rectangular to polar, we have

${(r^2=x^2+y^2), (tan(theta) = y/x):}$

Using the second pair of equations with $x=5$ and $y=-12$ , we have

$r^2 = 5^2+(-12)^2 = 169$

$=> r = sqrt(169) = 13$

and

$tan(theta) = -12/5$

$=> theta = arctan(-12/5) ~~ -67.380^@$

Thus, the polar coordinates for $(x, y) = (5, -12)$ are $(r, theta) = (13, arctan(-12/5)) ~~ (13, -67.380^@)$