# How do you find the polar coordinates given the rectangular coordinates (-1/4, -sqrt3/4)?

Aug 23, 2017

See explanation...

#### Explanation:

You can plot $- \frac{1}{4} , - \sqrt{\frac{3}{4}}$ on the Cartesian plane easily enough. We seek a representation of this same point in (2d) space using the polar coordinates $\left(r , \theta\right)$

The r is easy enough:

r = sqrt((-1/4)^2 + (-sqrt(3/4))^2

$= \sqrt{\frac{1}{16} + \frac{3}{4}}$

$= \frac{\sqrt{13}}{4}$

The value we calculate for the angle $\theta$ will depend on where we choose the $\theta = 0$ direction to be. This is an arbitrary choice. Here I will choose $\theta = 0$ to be on the x-axis to the right. So, therefore, $\theta = \frac{\pi}{2}$ will be on the y-axis pointing upwards, etc.

If $r = \frac{\sqrt{13}}{4}$, then $\frac{- \frac{1}{4}}{\frac{\sqrt{13}}{4}} = \cos \left(\theta\right)$

so therefore $\theta = \arccos \left(- \frac{1}{\sqrt{13}}\right)$

...but note that you can choose the $\theta = 0$ direction to be on the y axis pointing up. In this case,

$\frac{- \frac{1}{4}}{\frac{\sqrt{13}}{4}} = \sin \left(\theta\right)$.

and therefore $\theta$ would be $\arcsin \left(- \frac{1}{\sqrt{13}}\right)$

...I'm guessing your teacher may have a preferred direction for the $\theta = 0$ direction, so use whichever of these answers is appropriate.

GOOD LUCK.