# What is the polar form of ( -1,16 )?

May 6, 2018

the polar cordinates $\left(r , \theta\right) = \left(\sqrt{257} , - 1.5084\right)$

#### Explanation:

the point $\left(- 1 , 16\right)$

$r = \sqrt{{x}^{2} + {y}^{2}} = \sqrt{{\left(- 1\right)}^{2} + {\left(16\right)}^{2}} = \sqrt{257}$

$\tan \theta = \frac{y}{x} = \frac{16}{-} 1 = - 16$

we will take arctan to each side

$\arctan \left(\tan \theta\right) = \arctan \left(- 16\right)$

$\theta = \arctan \left(- 16\right)$

$\theta = - 1.5084$

the polar cordinates $\left(r , \theta\right) = \left(\sqrt{257} , - 1.5084\right)$