I want to know how to calculate $\theta$, every other variable in the picture is known. The solution should still work if there are more than $2$ triangles in the square. $a$ is always the same and I want to be able to change $r$ however I want.
( I was able to do it for $a = r$, $\theta = {\omega \over 2}$ with $\omega = 180 - \gamma$. But I couldn't figure it out for the general case. Maybe $\alpha$ (orange) looks a little bit like $\gamma$ (purple) in the picture, they are of course not the same, sorry about that.
You can create a right triangle by splitting $d$ in half and connecting this point to the center of the circle, I don't know if that helps but I used that triangle to calculate $\gamma$ with $\cos$. Other formulas I used are $l = r\alpha$ for the length of a circle piece and $s = 2r\sin{\alpha \over 2}$ for the distance of $2$ points on the edge of the circle, also don't know if that helps at all. )
I want to program something and this is the last piece I need, I really appreciate any help.
Notice that $\angle ACG=\angle AGC=\dfrac{1}{2}\left(\pi-\dfrac{\alpha}{2}\right)$.
Therefore $\angle CGE=\pi-\angle AGC=\dfrac{\pi}{2}+\dfrac{\alpha}{4}$.
$\angle GEC=\dfrac{\pi}{4}$ since the diagonal of a square bisects the angle.
Your angle $\theta=\angle ECG=\pi-\left(\dfrac{\pi}{4}+\dfrac{\pi}{2}+\dfrac{\alpha}{4}\right)=\dfrac{\pi-\alpha}{4}$