# How do you convert (2sqrt3, 2) into polar form?

Dec 18, 2016

$\left(4 , \frac{\pi}{6}\right)$

#### Explanation:

To convert from $\textcolor{b l u e}{\text{cartesian to polar form}}$

That is $\left(x , y\right) \to \left(r , \theta\right)$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{r = \sqrt{{x}^{2} + {y}^{2}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

and $\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{here " x=2sqrt3" and } y = 2$

$\Rightarrow r = \sqrt{{\left(2 \sqrt{3}\right)}^{2} + {2}^{2}} = \sqrt{12 + 4} = 4$

$\text{Now, " (2sqrt3,2)" is in the first quadrant }$so we must ensure that $\theta$ is in the first quadrant.

$\theta = {\tan}^{-} 1 \left(\frac{2}{2 \sqrt{3}}\right) = {\tan}^{-} 1 \left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$

and $\frac{\pi}{6} \text{ is the reference angle in the first quadrant}$

$\Rightarrow \left(2 \sqrt{3} , 2\right) \to \left(4 , \frac{\pi}{6}\right)$