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A hyperbolic triangle is determined up to isometry by its three angles. Indeed, the angle form of the hyperbolic law of cosines determines every side length from the angles, after which hyperbolic side-side-side congruence determines the triangle. This differs from Euclidean geometry because hyperbolic space has an intrinsic length scale.

Two corresponding side lengths and an equal nonincluded angle need not determine a hyperbolic triangle. Set $A=\pi /6$, $\cosh a=3/2$, $\cosh b=2$. The hyperbolic law of cosines has two distinct positive solutions $c=\ln (3\pm \sqrt{2})$. The two triangles share $a,b,A$ but have different third sides. An included angle instead determines the third side by the cosine rule.

In the hyperbolic law of cosines, the angle opposite $a$ being at least $\pi /2$ makes its cosine nonpositive. ThusSince $\cosh$ is increasing on $[0,\infty )$, the opposite side is longest: $\boxed{a\ge b,a\ge c}$. For a nondegenerate hyperbolic triangle both comparisons are strict.

In the curvature $-1$ normalization, the hyperbolic triangle area is its angle defect:Join one vertex of a convex geodesic polygon to its nonadjacent vertices. This gives $n-2$ hyperbolic triangles with disjoint interiors. Their angles at every polygon vertex sum to that vertex's interior angle. Adding the hyperbolic triangle areas therefore gives the hyperbolic polygon areaFor a regular hyperbolic polygon with prescribed area, place $n$ equally spaced vertices on a hyperbolic circle of radius $R>0$ and join consecutive vertices by geodesics. Rotations through $2\pi /n$ and reflections in radial lines show that this is a convex regular polygon. The triangle from its centre to a vertex and the midpoint of an adjacent side is right angled, with angles $\pi /n$, $\alpha (R)/2$, and $\pi /2$, and hypotenuse $R$. The angle form of the hyperbolic law of cosines givesConsequently $A(R)=(n-2)\pi -n\alpha (R)$ is continuous and strictly increasing. As $R\downarrow 0$, $\alpha (R)\to \pi -2\pi /n$, so $A(R)\to 0$; as $R\to \infty$, $\alpha (R)\to 0$, so $A(R)\to (n-2)\pi$. The intermediate value theorem proves existence for every permitted $A$. Indeed the required radius isThe argument also proves uniqueness of the radius in this construction; neither endpoint area is attained by a nondegenerate finite-radius polygon.

In curvature $-1$, a regular $n$-gon of circumradius $R>0$ has interior angle $\alpha (R)=2\arctan (\cot (\pi /n)/\cosh R)$, by the right-triangle angle form of the hyperbolic law of cosines. Thus its hyperbolic polygon area is $(n-2)\pi -n\alpha (R)$, continuously and strictly increasing from $0$ to $(n-2)\pi$. Every area in that open interval is therefore attained at exactly one circumradius.