# How do you convert (sqrt(3), 1) into polar forms?

Jan 9, 2016

If $\left(a , b\right)$ is a are the coordinates of a point in Cartesian Plane, $u$ is its magnitude and $\alpha$ is its angle then $\left(a , b\right)$ in Polar Form is written as $\left(u , \alpha\right)$.
Magnitude of a cartesian coordinates $\left(a , b\right)$ is given by$\sqrt{{a}^{2} + {b}^{2}}$ and its angle is given by ${\tan}^{-} 1 \left(\frac{b}{a}\right)$

Let $r$ be the magnitude of $\left(\sqrt{3} , 1\right)$ and $\theta$ be its angle.
Magnitude of $\left(\sqrt{3} , 1\right) = \sqrt{{\left(\sqrt{3}\right)}^{2} + {1}^{2}} = \sqrt{3 + 1} = \sqrt{4} = 2 = r$
Angle of $\left(\sqrt{3} , 1\right) = T a {n}^{-} 1 \left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$

$\implies$ Angle of $\left(\sqrt{3} , 1\right) = \frac{\pi}{6} = \theta$

$\implies \left(\sqrt{3} , 1\right) = \left(r , \theta\right) = \left(2 , \frac{\pi}{6}\right)$
$\implies \left(\sqrt{3} , 1\right) = \left(2 , \frac{\pi}{6}\right)$
Note that the angle is given in radian measure.