Calculate the angle between a side of a quadrilateral and its diagonal Given two adjacent sides and all four angles of a quadrilateral, what is the most efficient way to calculate the angle that is made between a side and the diagonal of the quadrilateral that crosses (but does not necessarily bisect) the angle in between the two known sides?
Other known information:


*

*The two angles that touch one and only one known side are right angles.

*The angle that touches both known sides equals $n-m$

*The angle that doesn't touch any known sides equals $180-n+m$, which can be inferred through the above statement and the rule that states that the interior angles of a quadrilateral must add up to $360$, although this is also known from other aspects of the broader problem

*$n$ and $m$ cannot easily be explained with words. See the picture at the end of this post.


From what I can tell, the most efficient solution to this problem is to solve for the OTHER diagonal using the law of cosines and the two known sides $x$ and $y$ from the sketch, use the law of sines and/or cosines to solve for the parts of angles $A$ and $C$ that make up the left-most triangle made by the other diagonal, find the other parts of angles $A$ and $C$ by using $90-A$ and $90-C$, respectively, since $A$ and $C$ are both right angles, then use the law of sines once more to find sides $AB$ and $BC$, and FINALLY use the law of sines to find any of the four thetas. Seems tedious. Am I missing something?
Here is an awful sketch I made of the problem:


Le us assume that $D$ lies at the origin and $\widehat{CDA}=\theta$. Then the coordinates of $A$ are $(x\cos\theta,x\sin\theta)$ and the line through $A$ which is orthogonal to $DA$ has equation $y(t)=-\frac{\cos\theta}{\sin\theta}t+\left(x\sin\theta+x\frac{\cos^2\theta}{\sin\theta}\right) $. It follows that the length of $CB$ is given by
$$ -\frac{\cos\theta}{\sin\theta} y + \left(x\sin\theta+x\frac{\cos^2\theta}{\sin\theta}\right)=\frac{x-y\cos\theta}{\sin\theta} $$
and by symmmetry the length of $AB$ is given by $\frac{y-x\cos\theta}{\sin\theta}$.
In equivalent terms: complete the quadrilateral, notice the relevant proportions then remove the attached pieces.
