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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.

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So, their is not a particular answer then? Their are many answers? – user74636 Apr 28 '13 at 1:51
For a quadrilateral with all equal interior angles? – user74636 Apr 28 '13 at 1:53
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
Wait, what do you mean Berci? – user74636 Apr 28 '13 at 2:11
He means something like this picture ($n=8$):… – Neal Apr 28 '13 at 3:06

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