= Solution
A <hyperbolic triangle> is bounded by three hyperbolic geodesic segments or rays. Its vertices may lie in the hyperbolic plane; an <ideal vertex> is their endpoint on the boundary at infinity. Every <ideal> angle is zero. <Gauss-Bonnet theorem> with curvature $-1$ gives
$$
\operatorname{area}(\Delta)=\pi-(\alpha+\beta+\gamma),
$$
so an all-ideal triangle has area $\pi$.
For fixed admissible angles, hyperbolic trigonometry determines all three side lengths from the angles, for example
$$
\cosh a=\frac{\cos\alpha+\cos\beta\cos\gamma}{\sin\beta\sin\gamma}.
$$
Thus two such triangles are congruent. An orientation-preserving isometry can send one chosen vertex and oriented tangent to the corresponding data of the other, and the determined side lengths and angles then send the entire triangle to it. Since the orientation-preserving isometry <group> is $PSL_2(\mathbb R)$, represented by $SL_2(\mathbb R)$, the stated action is transitive.
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