# Polar coordinates of a point are given. Find therectangular coordinates of the point. (-3,-135^(circ))

Polar coordinates of a point are given. Find therectangular coordinates of the point. $\left(-3,-{135}^{\circ }\right)$
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sviudes7w
That's not what I got.... I figured that since we're working with apoint in quadrant III, then the reference angle is 45 degrees,making it a $1-1-\sqrt{2}$ triangle.
Since $\mathrm{cos}\left(45\right)=\frac{x}{-3}$, then $x=-3\mathrm{cos}\left(45\right)=\frac{-3}{\sqrt{2}}$
By the same means, y equals the same value (since sin 45 and cos 45 are equal)
So the point is ($\frac{-3}{\sqrt{2}},\frac{-3}{\sqrt{2}}$)
###### Not exactly what you’re looking for?
Cristofer Graves
Given the polar coordinates: (r, $\theta$) = (-3, -135degrees)
Convert degrees to radians: (r, $\theta$) = (-3, -3$\pi$/4radians)
Use the following conversion:
$x=r\mathrm{cos}\theta$
$x=-3\mathrm{cos}\left(-3\pi /4\right)$
$x=\frac{-3}{1}\ast \frac{-\sqrt{2}}{2}=\frac{3\sqrt{2}}{2}$
$y=r\mathrm{sin}\theta$
$y=-3\mathrm{sin}\left(-3\pi /4\right)$
$y=\frac{-3}{1}\ast \frac{-\sqrt{2}}{2}=\frac{3\sqrt{2}}{2}$
Rectangular Coordinates are $\left(x,y\right)⇒\left(\frac{3\sqrt{2}}{2},\frac{3\sqrt{2}}{2}\right)$