Finding Angle of Irregular Quadrilateral I have a linkage with 4 vertices and 3 bars. One point is at origin and one point is at point X,Y. The lengths of the 3 bars are known. In order to keep the solutions less than infinite, I assume the two angles are equal. 
Given this info how can I solve for A?

Let $O$ be the origin, $A,B$ be the vertices with angel $A$, and $C$ be the point $(X,Y)$. We know $OC=\sqrt{X^2+Y^2}$. Let $OA$ intersect $CB$ at $D$. Then, $DAB$ is isosceles with base angles equal to $\pi-A$. Thus, $DA=DB=\dfrac a{-2\cos A}$. Also, using Cosine's Law on $DOC$, we have 
$$X^2+Y^2=OC^2=(a-\frac a{2\cos A})^2+(b-\frac a{2\cos A})^2+2(\cos 2A)(a-\frac a{2\cos A})(b-\frac a{2\cos A})$$
You need to solve this for $\cos A$, by plugging $\cos 2A=2\cos^2A-1$. However, this looks tedious. A better alternative is first drawing $OB$ and noticing $OAB$ is isosceles. Thus, the angle $OBC$ is $\frac32 A-\frac \pi2$, and $OB=2a\sin \frac A2$. Using Cosine's Law on $OBC$ gives
$$X^2+Y^2=4a^2\sin^2 \frac A2+b^2-4ab\sin\frac A2\sin\frac32A$$
, which is an equation of $\sin\frac A2$.
