= Solution
\b[Yes.] By an element of $PSL_2(\mathbb R)$, part (a) puts the side between the $\alpha$- and $\beta$-vertices on $\ell_+$ and the first vertex at $i$. A reflection if needed puts the triangle on the chosen side. The remaining freedom is the ratio $t>1$ of the heights of the second and first vertices, and the equation from part (c) determines it uniquely because $(t+t^{-1})/2$ is strictly increasing on $(1,\infty)$. The two incident circles are then fixed, so their other intersection and the triangle are fixed. Thus triangles with the same three angles are isometric, which is <Hyperbolic AAA congruence>.
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