# What is the polar form of ( -51,6 )?

##### 1 Answer
Aug 7, 2016

Polar form coordinates are $\left(3 \sqrt{293} , {173.29}^{o}\right)$

#### Explanation:

A rectangular coordinate $\left(x , y\right)$ can be written as polar coordinate $\left(r , \theta\right)$, where $x = r \cos \theta$, $y = r \sin \theta$, $\theta = {\tan}^{- 1} \left(\frac{y}{x}\right)$ and $r = \sqrt{{x}^{2} + {y}^{2}}$.

As we have rectangular coordinate $\left(- 51 , 6\right)$

$r = \sqrt{{\left(- 51\right)}^{2} + {6}^{2}} = \sqrt{2601 + 36} s q r 2637 = 3 \sqrt{293}$

and $\theta = {\tan}^{- 1} \left(- \frac{6}{51}\right) = {\tan}^{- 1} \left(- 0.11765\right) = {\left(180 - 6.71\right)}^{o} = {173.29}^{o}$

Note - As $\tan \theta$ is negative and while $\cos \theta$ is negative and $\sin \theta$ is positive, we have taken $\theta$ in second quadrant.

Hence polar form coordinates are $\left(3 \sqrt{293} , {173.29}^{o}\right)$