# What is the Cartesian form of (99,(17pi )/12)?

Mar 19, 2017

$\left(- 95.63 , - 25.62\right)$

#### Explanation:

polar coordinates $\left(r , \theta\right) = \left(99 , \frac{17 \pi}{12}\right)$ is in quadrant IV

rectangular coordinates (rsin theta, r cos theta)

$99 \sin \left(17 \frac{\pi}{12}\right) \approx - 95.6266568$

$99 \cos \left(17 \frac{\pi}{12}\right) \approx - 25.62308547$

Rectangular coordinates: $\left(- 95.63 , - 25.62\right)$

Mar 19, 2017

$- \frac{99 \left(\sqrt{3} - 1\right)}{2 \sqrt{2}} , - \frac{99 \left(\sqrt{3} + 1\right)}{2 \sqrt{2}}$

#### Explanation:

In polar coordinates it is $r = 99 \mathmr{and} \theta = \frac{17 \pi}{12}$. In cartesean form it would be $x = r \cos \theta$ and $y = r \sin \theta$

Accordingly cartesean coordinates would be $x = 99 \cos \left(\frac{17 \pi}{12}\right) = 99 \cos \left(\pi + \frac{5 \pi}{12}\right) = - 99 \cos \left(\frac{5 \pi}{12}\right) = - 99 \cos {75}^{o} = - \frac{99 \left(\sqrt{3} - 1\right)}{2 \sqrt{2}}$

similarly,$y = 99 \sin \left(\frac{17 \pi}{12}\right) = - 99 \sin \left(\frac{5 \pi}{12}\right) = - 99 \sin {75}^{o} = - \frac{99 \left(\sqrt{3} + 1\right)}{2 \sqrt{2}}$