## Precalculus (6th Edition) Blitzer

$(13, 67.4^{\circ})$
Here, $r=\sqrt {x^2+y^2}=\sqrt{(5)^2+(12)^2}= \sqrt {169}=13$ $\tan \theta =\dfrac{y}{x}$ and $\theta=arctan[\dfrac{12}{5}]$ Thus, $\theta =67.4^{\circ}$ Hence, $(13, 67.4^{\circ})$