# What is the polar form of ( 13,-4 )?

Nov 18, 2017

Polar form is $\left(13.6 , 5.985\right) \mathmr{and} \left(13.6 , - 0.298\right)$

#### Explanation:

$\left(13 , - 4\right)$ lies on $4$ th quadrant.

$r = \sqrt{{13}^{2} + {\left(- 4\right)}^{2}} = \sqrt{185} = 13.6$

$\tan \alpha = \frac{4}{13} \therefore \alpha = {\tan}^{-} 1 \left(\frac{4}{13}\right) = 0.298498$

Since $\theta$ lies $4$ th quadrant.

$\theta = 2 \pi - \alpha = 2 \pi - 0.298498 \approx 5.984686$ or

$\theta = \left(- \alpha\right) = - 0.298498$

Polar form is $\left(r , \theta\right) \therefore \left(13.6 , 5.985\right) \mathmr{and} \left(13.6 , - 0.298\right)$ [Ans]

Nov 18, 2017

$\left(\sqrt{185} , - \frac{19 \pi}{200}\right)$

#### Explanation:

Polar coordinate form is : $\left(r , \theta\right)$

Cartesian coordinate form $\left(x , y\right)$

$x = r \cos \theta$

$y = r \sin \theta$

$\theta = \arctan \left(\frac{y}{x}\right)$

$\therefore$

$13 = r \cos \theta \textcolor{w h i t e}{88} \left[1\right]$

$- 4 = r \sin \theta \textcolor{w h i t e}{88} \left[2\right]$ Squaring:

$169 = {r}^{2} {\cos}^{2} \theta$

$16 = {r}^{2} {\sin}^{2} \theta \textcolor{w h i t e}{88}$ adding [1] and [2]

$185 = {r}^{2} \left({\sin}^{2} \theta + {\cos}^{2} \theta\right)$$\textcolor{w h i t e}{88888} {\sin}^{2} \theta + {\cos}^{2} \theta = 1$

$\therefore$

$185 = {r}^{2} \implies r = \sqrt{185}$

$\tan \theta = - \frac{4}{13} \implies \theta = {\tan}^{-} 1 \left(- \frac{4}{13}\right) \approx - 0.2985 \approx - \frac{19 \pi}{200}$

$\therefore$

$\left(\sqrt{185} , - \frac{19 \pi}{200}\right)$