Is there any equation looks like Cagnolli’s first formula(for hyperbolic triangle ABC): $$\sin\frac{S}{2}=\frac{\sinh\frac{a}{2}\sinh\frac{b}{2}\sin C}{\cosh\frac{c}{2}}$$ for hyperbolic right-angled hexagons?
Or any formula between its area($\pi$)and lengths?
Thanks:)
The formula is:
$S = \pi,$ thanks to Gauss-Bonnet.
As Igor points out $S=\pi$ for a right-angled hexagon. In general, the area of a hyperbolic right-angled $n$-gon will be $\frac{\pi}{2}(n-4)$ (here $n>4$ is required for hyperbolicity). More generally, knowing the angles (in fact just the angle sum) and the number of sides will determine the area of a hyperbolic polygon. The formula in the question is equivalent to determining the sum of the two other angles in the triangle.
However, it may be worth mentioning that Chapter 2 of Thurston's notes has a treatment of hyperbolic right-angled polygons, which ends with a Law of Sines for right angled hexagons:
$$\frac{\sinh A}{\sinh \alpha}=\frac{\sinh B}{\sinh \beta}=\frac{\sinh C}{\sinh \gamma}$$
where $A,B,C, \alpha, \beta, \gamma$ are as in the figure below (taken from Thurston's notes Chapter 2, page 26).
