# How do you find the polar coordinate of (2sqrt3, 2)?

May 5, 2018

$\left(4 , \frac{\pi}{6}\right)$ (radians) or $\left(4 , {60}^{\circ}\right)$ (degrees)

#### Explanation:

Rectangular $\to$ Polar: $\left(x , y\right) \to \left(r , \theta\right)$

• Find $r$ (radius) using $r = \sqrt{{x}^{2} + {y}^{2}}$
• Find $\theta$ by finding the reference angle: $\tan \theta = \frac{y}{x}$ and use this to find the angle in the correct quadrant

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

$r = \sqrt{\left(4 \cdot 3\right) + 4}$

$r = \sqrt{12 + 4}$

$r = \sqrt{16}$

$r = 4$

Now we find the value of $\theta$ using $\tan \theta = \frac{y}{x}$.

$\tan \theta = \frac{2}{2 \sqrt{3}}$

$\tan \theta = \frac{1}{\sqrt{3}}$

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

$\theta = \frac{\pi}{6}$ or $\frac{7 \pi}{6}$

To determine which one it is, we have to look at our coordinate $\left(2 \sqrt{3} , 2\right)$. First, let's graph it:

As you can see, it is in the first quadrant. Our $\theta$ has to match that quadrant, meaning that $\theta = \frac{\pi}{6}$.

From $r$ and $\theta$, we can write our polar coordinate:
$\left(4 , \frac{\pi}{6}\right)$ (radians) or $\left(4 , {60}^{\circ}\right)$ (degrees)

Hope this helps!