In the above diagram, I am trying to solve for the 3 angles, x, y, and z (hope they are clearly labeled). The circles next to the angles just denote where the joints are; this is for a robot arm. Any solution of these angles will work, with the one caveat that none of the joints can go "above" the hypotenuse c. Distances a, b, and c are all known, as are the lengths of the three arm parts, h1, h2, and h3. h3 > h2 = h1 in length.
One thing that may help is that I can "fix" part h3 at a certain angle if need be, like saying that it will always be 5 degrees away from c. I think that may help, but I am not sure how. I think, if possible, it would actually be preferable to fix part h3. How can I solve for x, y, and z in terms of the other knowns? Thanks in advance for your help, I am a trigonometry/inverse kinematics noob, and Google has not led to a solution so far.
In most cases there is no unique solution. For example, say $c/3<h_1=h_2=h_3<c$. I can show 3 trivial solutions:
But there are infinitely many possibilities, so you need more constraints.
EDIT For case 3, you know that the long base of the trapezoid is $c$, the short base is $h=h_1=h_2=h_3$. If you draw perpendiculars from the $y$ and $z$ joints onto $c$ you form a rectangle and two right angle triangles. The hypotenuse if $h$ and one of the sides is $$\frac{c-h}2$$Then the angle between $c$ and $h_1$ is $$\alpha=\arccos\frac{c-h}{2h}$$ The angle between $h_1$ and $h_2$, and the angle between $h_2$ and $h_3$ will be $$90^\circ+(90^\circ-\alpha)=180^\circ-\alpha$$
EDIT2 New request: $h_1=h_2=h<h_3$. We are still going to try to create a trapezoid, $h_2||c$, but it won't be isosceles anymore. In your sketch, draw a parallel to $h_3$ through the joint between $h_1$ and $h_2$. You will have a parallelogram with lengths $h_3$ and $h$, and a triangle with lengths $h$, $h_3$, and $c-h$. You can use Law of cosines to calculate the angles in this triangle.