# What is the polar form of ( 10,10 )?

Feb 20, 2016

Use the following formulas:

$r = \sqrt{{x}^{2} + {y}^{2}}$

$\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right)$

#### Explanation:

The rectangular coordinate $\left(10 , 10\right)$ lies in quadrant I. So, $\theta$ must fall between $0$ and $90$ degrees

First, solve for r:

$r = \sqrt{{x}^{2} + {y}^{2}} = \sqrt{{10}^{2} + {10}^{2}} = \sqrt{2 \times 100} = 10 \sqrt{2}$

Now, find $\theta$ ...

$\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right) = {\tan}^{-} 1 \left(\frac{10}{10}\right) = {\tan}^{-} 1 \left(1\right) = {45}^{o}$ or $\frac{\pi}{4}$

Polar Form : $\left(r , \theta\right) = \left(10 \sqrt{2} , \frac{\pi}{4}\right)$

hope that helped