Since you have the lengths and directions of two sides of the triangle, you can compute the coordinates of its right-hand vertex relative to the centre of the circle; their ratio gives you $\tan\alpha$:
$$ \alpha=\arctan\frac{r\sin\partial+d\sin(\partial-\pi+L)}{r\cos\partial+d\cos(\partial-\pi+L)}\;, $$
where $r$ is the radius and $d$ is the length of the red line.