Rotate Object About a Point With Crank Mechanism I am trying to figure out how to create this physics simulation but I need some guidance on how to go about calculating it. Below is an image of the system I am working to solve.

Here I am trying to find out how to calculate the angle at which to rotate the black triangle in order to maintain the length of 111 for the blue linkage as the circle crank rotates continuously. I have used the law of cosines to calculate all the angles and the changing length (x). What I am stuck on is how to take all the variables and put them together to calculate the proper triangle rotation to maintain the 111 length. 
Any help would be greatly appreciated and if I can give any more information please let me know. 
 A: You were right to think of the Law of Cosines in this problem.
We will need to use the center point of the circle for reference. Since you have not labeled it, I will call that point $d$.
The first step is to find $x$, which is given by the Pythagorean Theorem
$$x=\sqrt{(94.017+38.4\cos\theta)^2+(38.4\sin\theta)^2}\approx\sqrt{10313.8+7220.51\cos(\theta)}$$
Where $\theta$ is the angle between $\overline{dc}$ and the rightward horizontal.
Next, use the Law of Cosines to solve for $\angle{abc}$.
$$111^2=100^2+x^2-2(100)x\dot{}\cos\angle{abc}\implies
\angle{abc}=\cos^{-1}\left(\frac{x^2-2321}{200x}\right)$$
Then use the Law of Sines to find $\angle{cbe}$. Notice that $\angle{bdc}=\pi-\theta$
$$\frac{\sin\angle{cbd}}{38.4}=\frac{\sin{(\pi-\theta)}}{x}\implies
\angle{cbd}=\sin^{-1}\left(\frac{38.4\sin{(\pi-\theta)}}{x}\right)$$
Finally, we find $\angle{abd}$ (the angle between $\overline{ba}$ and the rightward horizontal)
$$\angle{abd}=\angle{abc}+\angle{cbd}=\cos^{-1}\left(\frac{x^2-2321}{200x}\right)+\sin^{-1}\left(\frac{38.4\sin{(\pi-\theta)}}{x}\right)$$
Solve the above equation numerically; substituting the expression for $x$ does not simplify nicely and will lead to rounding errors.
Knowing $\angle{abd}$ should be sufficient information to translate the other points in the image.
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