TITLE: How to find the vertices angle after rotation
QUESTION [18 upvotes]: I drew this picture to interpret my question.

I have the x and y axis for all the vertices before rotating the object. And I have the angle of rotation, how can I find the x and y axis for the vertices after rotating the object.
Thanks

REPLY [25 votes]: When a point $(x,y)$ is rotated about the origin $(0,0)$ counterclockwise by an angle $\theta$, the coordinates of the new point $(x',y')$ are
$$\begin{align} x'&=x\cos(\theta)-y\sin(\theta), \\ y'&=x\sin(\theta)+y\cos(\theta).\end{align}$$
Thus, when we rotate a point $(x,y)$ about another point $(p,q)$ counterclockwise by an angle $\theta$, we can compute the new point's coordinates by 

translating the entire plane so that $(p,q)$ goes to the origin,
perform the rotation, and then
translate the entire plane back.

To translate $(p,q)$ to the origin, we subtract $p$ from $x$-coordinates and $q$ from $y$-coordinates, and to undo the operation we add instead of subtract. Thus, for example, after translating $(p,q)$ to the origin, the coordinates $(x,y)$ of our point have become $(x-p,y-q)$.
Therefore, the new point's coordinates are 
$$\begin{align} x'&=(x-p)\cos(\theta)-(y-q)\sin(\theta)+p, \\ y'&=(x-p)\sin(\theta)+(y-q)\cos(\theta)+q.\end{align}$$
In your particular case, we can now see that the coordinates of the points $a$, $b$, $c$, and $d$ after rotation are 
$$\begin{align}
a'&=((1-3)\cos(15)-(5-3)\sin(15)+3,(1-3)\sin(15)+(5-3)\cos(15)+3)\\\\
&=\left((-2)\left(\tfrac{1+\sqrt{3}}{2\sqrt{2}}\right)-(2)\left(\tfrac{-1+\sqrt{3}}{2\sqrt{2}}\right)+3,(-2)\left(\tfrac{-1+\sqrt{3}}{2\sqrt{2}}\right)-(2)\left(\tfrac{1+\sqrt{3}}{2\sqrt{2}}\right)+3\right)\\\\
&=(3-\sqrt{6},3+\sqrt{2})\\\\
&\approx(0.55051,4.41421)\\\\\\
b'&=((5-3)\cos(15)-(5-3)\sin(15)+3,(5-3)\sin(15)+(5-3)\cos(15)+3)\\\\
&=\left((2)\left(\tfrac{1+\sqrt{3}}{2\sqrt{2}}\right)-(2)\left(\tfrac{-1+\sqrt{3}}{2\sqrt{2}}\right)+3,(2)\left(\tfrac{-1+\sqrt{3}}{2\sqrt{2}}\right)-(2)\left(\tfrac{1+\sqrt{3}}{2\sqrt{2}}\right)+3\right)\\\\
&=(3+\sqrt{2},3+\sqrt{6})\\\\
&\approx(4.41421,5.44949)\\\\\\
c'&=((1-3)\cos(15)-(1-3)\sin(15)+3,(1-3)\sin(15)+(1-3)\cos(15)+3)\\\\
&=\left((-2)\left(\tfrac{1+\sqrt{3}}{2\sqrt{2}}\right)-(-2)\left(\tfrac{-1+\sqrt{3}}{2\sqrt{2}}\right)+3,(-2)\left(\tfrac{-1+\sqrt{3}}{2\sqrt{2}}\right)-(-2)\left(\tfrac{1+\sqrt{3}}{2\sqrt{2}}\right)+3\right)\\\\
&=(3-\sqrt{2},3-\sqrt{6})\\\\
&\approx(1.58579,0.55051)\\\\\\
d'&=((5-3)\cos(15)-(1-3)\sin(15)+3,(5-3)\sin(15)+(1-3)\cos(15)+3)\\\\
&=\left((2)\left(\tfrac{1+\sqrt{3}}{2\sqrt{2}}\right)-(-2)\left(\tfrac{-1+\sqrt{3}}{2\sqrt{2}}\right)+3,(2)\left(\tfrac{-1+\sqrt{3}}{2\sqrt{2}}\right)-(-2)\left(\tfrac{1+\sqrt{3}}{2\sqrt{2}}\right)+3\right)\\\\
&=(3+\sqrt{6},3-\sqrt{2})\\\\
&\approx(5.44949,1.58579)\\\\\\
 \end{align}$$

Plotting these points in Mathematica demonstrates visually that our calculations were correct:
ListPlot[{{3 - Sqrt[6], 3 + Sqrt[2]}, {3 + Sqrt[2], 3 + Sqrt[6]}, 
{3 - Sqrt[2], 3 - Sqrt[6]}, {3 + Sqrt[6], 3 - Sqrt[2]}}, AspectRatio -> 1, 
AxesOrigin -> {0, 0}, PlotMarkers -> {Automatic, Medium}, PlotStyle -> Blue]