# How do you find the polar coordinate of the following point (-2, 5)?

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 $\left(- 2 , 5\right)$ and $\theta$ be its angle.
Magnitude of $\left(- 2 , 5\right) = \sqrt{{\left(- 2\right)}^{2} + {5}^{2}} = \sqrt{4 + 25} = \sqrt{29} = r$
Angle of $\left(- 2 , 5\right) = T a {n}^{-} 1 \left(\frac{5}{- 2}\right) = T a {n}^{-} 1 \left(- \frac{5}{2}\right) = - 68.198$ degree

$\implies$ Angle of $\left(- 2 , 5\right) = - 68.198$ degree

But since the point is in second quadrant so we have to add $180$ degree which will give us the angle.

$\implies$ Angle of $\left(- 2 , 5\right) = - 68.198 + 180 = 111.802$

$\implies$ Angle of $\left(- 2 , 5\right) = 111.802 = \theta$

$\implies \left(- 2 , 5\right) = \left(r , \theta\right) = \left(\sqrt{29} , 111.802\right)$
$\implies \left(- 2 , 5\right) = \left(\sqrt{29} , 111.802\right)$
Note that the angle is given in degree measure.