# What is the polar form of ( -3,99 )?

$\left(99.05 , 1.6\right)$
$r = \sqrt{{x}^{2} + {y}^{2}} = \sqrt{{\left(- 3\right)}^{2} + {\left(99\right)}^{2}} \approx 99.05$
$\theta = {\tan}^{-} 1 \left(\frac{99}{-} 3\right) + \pi \approx 1.6$
$\left(99.05 , 1.6\right)$