Solve $d_h(A,B)$ on a Poincare Disc

Question

Consider △ABC on a poincare disc.

On △ABC,

$\angle C = \theta(radian)$, $d_h(B,C)=d_h(A,C)=b$

In this situation, solve $d_h(A,B)$.

To me, it is hard because I have no experience.

Is there someone to help?

Answer 1

In hyperbolic geometry, there are two laws of cosine, as opposed to the one in Euclidean geometry. One relates three lengths and one angle (like in the Euclidean case), the other three angles and one length (which wouldn't work in Euclidean geometry due to similarities). You need the one for three lengths and one angle, just like in a Euclidean triangle. The formula is commonly written like this:

$$\cosh c = \cosh a \cosh b - \sinh a \sinh b \cos \gamma$$

Since $\gamma$ corresponds to your angle $\theta$, the two lengths you have correspond to $a$ and $b$, and the length you want to compute is $c$. You can use the fact that $a=b$ in your case to simplify the formula using squares.