I have to find $\epsilon$ and $\delta$.
The other angles defined are known. The white line is split in half by the dashed line.
I checked in a CAD-Program that $\alpha$, $\beta$ and $\gamma$ are sufficient to define the shape. I am able to calculate all angles in the diagram except for $\epsilon$ and $\delta$. Only thing I could come up with was a relation between those:
$\epsilon + \delta =180^\circ- \alpha - \beta$
To solve for $\epsilon$ and $\delta$, I know only need another equation that is not redundant with the first. It needs to include $\gamma$, but I couldn't find such a formula. It seems like an easy trigonometry problem, but every approach I tried just resolves in a redundant formula.
Thank you for your help.
PS: If someone wants to know. I need those angles to calculate the gapsize of a leaf seal as a function of the radius.
Repeatedly applying sine law gives the two sides of the triangle at the bottom to be
$\frac{\sin{\alpha}\sin{(\pi/2+\gamma/2-\alpha-\beta)}}{\sin{(\alpha+\beta)}\sin\beta}$ for the side opposite to $\delta$, and
$\frac{\sin(\alpha+\beta-\gamma)}{2\sin(\gamma/2)\sin(\alpha+\beta)}$ for the side opposite to $\epsilon=\pi-\alpha-\beta-\delta$.
Now use sine law on this triangle, $$\frac{\sin{\alpha}\sin{(\pi/2+\gamma/2-\alpha-\beta)}}{\sin{(\alpha+\beta)}\sin\beta\sin\delta} = \frac{\sin(\alpha+\beta-\gamma)}{2\sin(\gamma/2)\sin(\alpha+\beta)\sin(\pi-\alpha-\beta-\delta)}$$
$\delta$ can now be solved to be $$\cot\delta=\frac{\sin\beta\sin(\alpha+\beta-\gamma)}{2\sin\alpha\sin(\gamma/2)\sin(\alpha+\beta)\cos(\alpha+\beta-\gamma/2)}-\cot(\alpha+\beta)$$