# How do you convert the Cartesian coordinates (2√3, 2) to polar coordinates?

Sep 5, 2015

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

#### Explanation:

Let $\left(x , y\right)$ be a coordinate on the Cartesian plane.

The corresponding polar coordinate is $\left(r , \theta\right)$, where:

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

(You might notice that this is similar to the distance formula; that's not a coincidence, $r$ is the distance from the point to the pole (a.k.a. the center) )

and:

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

So, given $\left(2 \sqrt{3} , 2\right)$:

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

$r = \sqrt{12 + 4}$

$r = \sqrt{16}$

$r = 4$

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

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

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

$\theta = \frac{\pi}{6}$

We can say that $\theta = \frac{\pi}{6}$ (and not $\frac{5 \pi}{6}$ , etc.) because $x$ and $y$ are both positive, which means the point is in the first quadrant.

Thus, the point in polar coordinates is $\left(4 , \frac{\pi}{6}\right)$.