# Question #00d88

Nov 29, 2016

$\left(r , \theta\right) = \left(13 , \arctan \left(- \frac{12}{5}\right)\right) \approx \left(13 , - {67.380}^{\circ}\right)$

#### Explanation:

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

$\left\{\begin{matrix}x = r \cos \left(\theta\right) \\ y = r \sin \left(\theta\right)\end{matrix}\right.$

and to go from rectangular to polar, we have

$\left\{\begin{matrix}{r}^{2} = {x}^{2} + {y}^{2} \\ \tan \left(\theta\right) = \frac{y}{x}\end{matrix}\right.$

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

${r}^{2} = {5}^{2} + {\left(- 12\right)}^{2} = 169$

$\implies r = \sqrt{169} = 13$

and

$\tan \left(\theta\right) = - \frac{12}{5}$

$\implies \theta = \arctan \left(- \frac{12}{5}\right) \approx - {67.380}^{\circ}$

Thus, the polar coordinates for $\left(x , y\right) = \left(5 , - 12\right)$ are $\left(r , \theta\right) = \left(13 , \arctan \left(- \frac{12}{5}\right)\right) \approx \left(13 , - {67.380}^{\circ}\right)$