On the hyperbolic plane, if I have a quadrilateral that has all congruent interior angles $\alpha$, how do I figure out what $\alpha$ is? I know in Euclidean geometry one could just use $\frac{180(n-2)}{n}$.

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Do you mean quadrilateral, that is specifically four sides, or polygon? You use of $n$ indicates you may be interested in more general polygons. I think the answer is the same, as given by Berci. – Ross Millikan Apr 28 '13 at 2:45

This can be any $\alpha<\displaystyle\frac{180^\circ(n-2)}n$. (For any such $\alpha$ there exists an $n$-gon with equal sides.)
For illustration, pick a regular $n$-gon, and zoom its vertices out from the center of the $n$-gon (equally raise the distance on the rays from the center), then connect again the new $n$ vertices. You will get a regular $n$-gon with bigger area and smaller angles.
Well, there is no regular zooming in hyperbolic plane (at least which preserves lines), but the vertices can be zoomed out and they will form a bigger/smaller regular $n$-gon with smaller/bigger angle (remember that the sum of the angles is in correlation with the area). – Berci Apr 28 '13 at 1:53
He means something like this picture ($n=8$): simonsfoundation.org/wp-content/uploads/2012/09/… – Neal Apr 28 '13 at 3:06