Polar coordinates of a point are given. Find therectangular coordinates of the point. $(-3,-{135}^{\circ})$

Lilliana Livingston
2022-07-27
Answered

Polar coordinates of a point are given. Find therectangular coordinates of the point. $(-3,-{135}^{\circ})$

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sviudes7w

Answered 2022-07-28
Author has **12** answers

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 $\mathrm{cos}(45)=\frac{x}{-3}$, then $x=-3\mathrm{cos}(45)=\frac{-3}{\sqrt{2}}$

By the same means, y equals the same value (since sin 45 and cos 45 are equal)

So the point is ($\frac{-3}{\sqrt{2}},\frac{-3}{\sqrt{2}}$)

Since $\mathrm{cos}(45)=\frac{x}{-3}$, then $x=-3\mathrm{cos}(45)=\frac{-3}{\sqrt{2}}$

By the same means, y equals the same value (since sin 45 and cos 45 are equal)

So the point is ($\frac{-3}{\sqrt{2}},\frac{-3}{\sqrt{2}}$)

Cristofer Graves

Answered 2022-07-29
Author has **6** answers

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 $

$x=-3\mathrm{cos}(-3\pi /4)$

$x=\frac{-3}{1}\ast \frac{-\sqrt{2}}{2}=\frac{3\sqrt{2}}{2}$

$y=r\mathrm{sin}\theta $

$y=-3\mathrm{sin}(-3\pi /4)$

$y=\frac{-3}{1}\ast \frac{-\sqrt{2}}{2}=\frac{3\sqrt{2}}{2}$

Rectangular Coordinates are $(x,y)\Rightarrow (\frac{3\sqrt{2}}{2},\frac{3\sqrt{2}}{2})$

Convert degrees to radians: (r, $\theta $) = (-3, -3$\pi $/4radians)

Use the following conversion:

$x=r\mathrm{cos}\theta $

$x=-3\mathrm{cos}(-3\pi /4)$

$x=\frac{-3}{1}\ast \frac{-\sqrt{2}}{2}=\frac{3\sqrt{2}}{2}$

$y=r\mathrm{sin}\theta $

$y=-3\mathrm{sin}(-3\pi /4)$

$y=\frac{-3}{1}\ast \frac{-\sqrt{2}}{2}=\frac{3\sqrt{2}}{2}$

Rectangular Coordinates are $(x,y)\Rightarrow (\frac{3\sqrt{2}}{2},\frac{3\sqrt{2}}{2})$

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