Calculated rotated point coordinate: is my solution correct 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
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!
Result of my calculation is slightly different than yours . 

Let's denote :

$\theta=-49.45^\circ $

$x&#39;=r\cos \alpha ~\text{ and }~y&#39;=r\sin \alpha$

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 : 

$(x&#39;,y&#39;)=(291.69,262.64)$