# What is the polar form of (-1,12)?

Mar 30, 2016

$\left(\sqrt{145} , {94.76}^{o}\right)$

#### Explanation:

x=-1, y=12;r = sqrt(x^2+y^2)=sqrt145.

$\theta$ satisfying both $\cos \theta = \frac{x}{r} = - \frac{1}{\sqrt{145}}$ and $\sin \theta = \frac{y}{r} = \frac{12}{\sqrt{145}}$ is

94,76^o#,
in the second quadrant in which cosine is negative and sine is positive.

$\sin \left({180}^{o} - \theta\right) = \sin \theta$. Calculator display for ${\sin}^{- 1} \left(\frac{12}{\sqrt{145}}\right)$ is ${85.24}^{o}$, nearly.
Choose $\theta = {180}^{o} - {85.24}^{o} = {94.76}^{o} = {\cos}^{- 1} \left(- \frac{1}{\sqrt{145}}\right)$.