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Is there a way to determine the angle measure of a regular polygon in hyperbolic space? I know that this depends on the length of the sides. A an example, I know that for an equilateral triangle with side length $a$ and angle $A$, then $$\sec A = 1 + \frac{2e^a}{e^{2a}+1}$$ Is there a similar formula for higher regular polygons?

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You can use the hyperbolic cosine rule for angles, namely $$ \cos{\alpha} = -\cos{\beta}\cos{\gamma} +\sin{\beta}\sin{\gamma}\cosh{a}, $$ on a triangle with vertices at the centre of the polygon and two consecutive vertices of the polygon. Taking $a$ as the side length, the central angle is obviously $2\pi/n$ and the triangle is isosceles, so $\beta=\gamma$, which gives $$ \cos{\tfrac{2\pi}{n}} = -\cos^2{\beta}+\sin^2{\beta}\cosh{a} = -\frac{1}{2}(1+\cos{2\beta}) + \frac{1}{2}(1-\cos{2\beta})\cosh{a}. $$ The internal angle $A$ is then $2\beta$, so we find after some hyperbolic identities that $$ \sec{A} = \tanh^2{\tfrac{1}{2}a} - \cos{\tfrac{2\pi}{n}}\operatorname{sech}^2{\tfrac{1}{2}a} $$

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