Your formulas are off.
You want the angles $36^\circ,\ 2\cdot36^\circ,\ 3\cdot36^\circ,\ \ldots,\ 10\cdot36^\circ.$
To convert degrees to radians, multiply by ${\pi\over 180}$.
So the angles in radians are ${36\pi\over 180} ,\ {2\cdot36\pi\over 180} ,\ {3\cdot36\pi\over 180} ,\ \ldots, {10\cdot36\pi\over 180} .$
With center $(50,50)$ and radius $r$, the point $(x,y)$ on the circle is $x=50+r \cos\theta, \quad y=50+r \sin\theta.$ where $r$ is the radius of the circle (you didn't specify $r$).
So for the 36 degree angle, for example, the point is $\textstyle x=50+r\cos( {36\pi\over 180} ),\quad y=50+r\sin( {36\pi\over 180}\cdot36 ).$ or, simplifying $\textstyle x=50+r\cos( { \pi\over5} ),\quad y=50+r\sin( { \pi\over5} ).$
(Note the angles are just ${\pi\over 5}$, ${2\pi\over 5}$, $\ldots\,$, ${10\pi\over 5}$.)
You could save some computational effort by taking advantage of symmetry:
The three points in the "first quadrant" of the circle are (corresponding to the angles ${10\pi\over5}=2\pi\sim0$, ${\pi\over 5}$, and ${2\pi\over5}$):
$\textstyle ( 50+r,50 ),\quad (50+r\cos{\pi\over5}, \quad 50+\sin{\pi\over5}),\quad (50+r\cos{2\pi\over5}, 50+\sin{2\pi\over5}) $
The three points in the "second quadrant" of the circle are (corresponding to the angles ${5\pi/5}=\pi $, ${3\pi\over 5}$, and ${4\pi\over5}$):
$\textstyle ( 50-r,50 ),\quad (50-r\cos{\pi\over5}, \quad 50+\sin{\pi\over5}),\quad (50-r\cos{2\pi\over5}, 50+\sin{2\pi\over5}) $
And to get the other four points in the "bottom half" of the circle, take the points above, except $(50+r,50)$ and $50-r,50)$, and switch the sign before the "$\sin$" in the $y$ coordinates.