# What is the polar form of ( 13,1 )?

Mar 22, 2018

$\left(\sqrt{170} , {\tan}^{-} 1 \left(\frac{1}{13}\right)\right) \equiv \left(13.0 , {0.0768}^{c}\right)$

#### Explanation:

For a given set of coordinates $\left(x , y\right)$, $\left(x , y\right) \to \left(r \cos \theta , r \sin \theta\right)$

$r = \sqrt{{x}^{2} + {y}^{2}}$
$\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right)$

$r = \sqrt{{13}^{2} + {1}^{2}} = \sqrt{169 + 1} = \sqrt{170} = 13.0$
$\theta = {\tan}^{-} 1 \left(\frac{1}{13}\right) = {0.0768}^{c}$

$\left(13 , 1\right) \to \left(\sqrt{170} , {\tan}^{-} 1 \left(\frac{1}{13}\right)\right) \equiv \left(13.0 , {0.0768}^{c}\right)$