# What is the polar form of (1,3)?

$\left(r , \theta\right) = \left(\sqrt{10} , \arctan \left(3\right)\right) \approx \left(3.16 , 1.25\right)$, where the second number is measured in radians.
The polar coordinates $\left(r , \theta\right)$ of a point with rectangular coordinates $\left(x , y\right)$ satisfy ${r}^{2} = {x}^{2} + {y}^{2}$ and $\tan \left(\theta\right) = \frac{y}{x}$ (when $x \ne 0$). Since the point $\left(x , y\right) = \left(1 , 3\right)$ is in the first quadrant, we can use the arctangent function to solve for the angle if we take $r$ to be the positive square root of ${x}^{2} + {y}^{2} = {1}^{2} + {3}^{2} = 10$.