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Is my calculation correct for this rotation around a point?

A point a(-19.94,392.11) is rotated -49.45 degrees, what is the new coordinates of point a?

My solution:

x' = x*cos(0) - y*sin(0)
y' = x*sin(0) + y*cos(0)

x' = (-12.961) - (-298.0036)
y' = (15.15) + (254.92)

x' = 285.04
y' = 270.07
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Is there a connection between this question and this other question posted a quarter of an hour later? It sure looks like it; even the $\theta$ written like a $0$ is the same. – joriki Mar 21 '12 at 10:55
@joriki sure is :P, I made this question also(before I remembered I had a registered account I could use) – Jake M Mar 21 '12 at 11:25

I don't know why this old point emerged... Anyway let's try this using complex numbers : $$(-19.94+392.11 i)\cdot e^{2\pi i\dfrac{-49.45}{360}}\approx 284.98 + 270.07i$$

So that the OP's answer looked not so bad!

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Result of my calculation is slightly different than yours .

Let's denote :

$\theta=-49.45^\circ $

$x'=r\cos \alpha ~\text{ and }~y'=r\sin \alpha$

where $~r=\sqrt{x^2+y^2} =392.5$

Note that :



hence :


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so is my answer incorrect? – Jake M Mar 21 '12 at 11:26
@JakeM Obviously..... – pedja Mar 21 '12 at 11:46

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