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

Oct 5, 2016

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

#### Explanation:

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

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

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

here $x = 2 \text{ and } y = 2 \sqrt{3}$

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

Now $\left(2 , 2 \sqrt{3}\right)$ is in the first quadrant so we must ensure that $\theta$ is in the first quadrant.

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

$= {\tan}^{-} 1 \left(\sqrt{3}\right) = \frac{\pi}{3} \leftarrow \text{ in first quadrant}$

$\Rightarrow \left(2 , 2 \sqrt{3}\right) \to \left(4 , \frac{\pi}{3}\right) \to \left(4 , {60}^{\circ}\right) \text{ in polar form}$