Exact Solution for Simple Sliding Linkage I am working with a linkage that consists of two cylinders with lengths $b$ and $c$ that actuate a platform to height $h$ and angle $\theta$.

The $h$ link is always perpendicular to the bottom two $a$ links which are fixed. The top two $a$ links are collinear. The value of $a$ is a known constant, and the values of $h$ and $\theta$ change based on the lengths of the cylinders. The dashed lines are shown as a reflection of the left side to illustrate the difference in angle between $b$ and $c$.
I have derived the formulas to calculate the exact lengths of the two cylinders $b$ and $c$ for given values of $h$ and $\theta$.
$$b^2 = {(h+a\sin\theta)}^2 + {(a-a\cos\theta)}^2$$
$$c^2 = {(h-a\sin\theta)}^2 + {(a-a\cos\theta)}^2$$
My problem is that I also want to solve it the other way, using values for $b$ and $c$ to find $h$ and $\theta$. From the above two equations, I derived the following:
$$h = \frac{b^2 - c^2}{4a\sin\theta} $$
I am currently using the approximation that $a\sin\theta=\frac{b-c}{2}$ which allows me to use the above equation to find $h$. This assumes that the very thin triangle on the right has no area, which is only actually true when the platform is horizontal.
The approximation is pretty close, but I would like to know the exact solution to calculate $h$ and $\theta$ from given values of $b$ and $c$.

I don't know that there is a "pretty" way to solve it symbolically. It is, however, technically possible to separate the variables and derive "exact" equations in either variable alone, which could then be solved numerically to the desired precision. One way to do that:
$$
b^2 - a^2(1-\cos\theta)^2 = {(h+a\sin\theta)}^2 \\
c^2 - a^2(1-\cos\theta)^2 = {(h-a\sin\theta)}^2
$$
Multiplying the above:
$$
\big(b^2 - a^2(1-\cos\theta)^2\big)\big(c^2 - a^2(1-\cos\theta)^2\big) = \big(h^2 - a^2 (1 - \cos^2 \theta)\big)^2 \tag{1}
$$
Using the other posted equation:

$$h = \frac{b^2 - c^2}{4a\sin\theta} \tag{2} \;\;\implies\;\; h^2 = \frac{(b^2-c^2)^2}{16a^2(1 - \cos^2 \theta)}$$

Substituting $h^2$ from $(2)$ back into $(1)$ gives in the end an $8^{th}$ degree polynomial equation in $\cos \theta$.
