Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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
share|improve this question
    
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

2 Answers 2

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!

share|improve this answer

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 :

$\alpha=\arctan\left|\frac{y}{x}\right|+\frac{\pi}{2}-|\theta|$

$\alpha=42.07^\circ$

hence :

$(x',y')=(291.69,262.64)$

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.