Edge length of hyperbolic tesselations If I have a general uniform tesselation in hyperbolic plane (same configuration of regular polygons at every vertex, but multiple types of polygons allowed), how can I find the edge length and/or inner angles of the polygons so they'd fit?
For example: a truncated {3,7} tesselation has two hexagon and one heptagon at every vertex. What is its edge length and what are the inner angles of the two types of polygons?
 A: The angle $\theta$ at a vertex of a regular hyperbolic $n$-gon with side length $s$ is given by
$$\sin\tfrac12\theta=\frac{\cos\tfrac12\tfrac{2\pi}n}{\cosh\tfrac12s}.$$
This is monotonically decreasing in $s$. As the polygon shrinks we have $\lim_{s\to0}\sin\tfrac12\theta=\cos\tfrac12\tfrac{2\pi}n$, or $\lim_{s\to0}\theta=\pi-\frac{2\pi}n$, the Euclidean $n$-gon angle. As the polygon grows we have $\lim_{s\to\infty}\theta=0$.
If your hyperbolic tiling has a vertex type $(n_1.n_2.n_3.\cdots)$, for the polygons to fit exactly around the vertex we need
$$2\pi=\sum_k\theta_k=\sum_k2\arcsin\frac{\cos\tfrac12\tfrac{2\pi}{n_k}}{\cosh\tfrac12s}.$$
This sum is also monotonically decreasing, so there is at most one solution $s$. There is a solution if and only if the sum in the Euclidean case $s\to0$ is greater than $2\pi$ (using the Intermediate Value Theorem). I don't think we can find the solution in closed form, in general.
Multiplying that equation by $\tfrac12i$ and exponentiating gives us this product of unit complex numbers:
$$-1=\prod_ke^{i\theta_k/2}=\prod_k\left(\sqrt{1-\left(\frac{\cos\tfrac12\tfrac{2\pi}{n_k}}{\cosh\tfrac12s}\right)^2}+i\frac{\cos\tfrac12\tfrac{2\pi}{n_k}}{\cosh\tfrac12s}\right);$$
since $\cos\tfrac\pi n$ is always an algebraic number, this shows that $\cosh\tfrac12s$ is also an algebraic number.
