Geometrical interpretation of the conjugacy of triangle groups. Let $\triangle$ and $\triangle'$  be two hyperbolic triangles of respective angles $\alpha,\beta,\gamma$ and $\alpha',\beta',\gamma'$. Let us suppose that the triangle subgroups $(\alpha,\beta,\gamma)$ and $(\alpha',\beta',\gamma')$ of $\mathbb{P}\mathrm{SL}(2,\mathbb{R})$ are conjugated. What does it tell us about the triangles $\triangle$ and $\triangle'$?
Reciprocally, is there any condition we could impose on the triangles $\triangle$ and $\triangle'$ for the triangle groups $(\alpha,\beta,\gamma)$ and $(\alpha',\beta',\gamma')$ to be conjugated?
 A: Assume that the angles of your triangles are all of the form $\pi/n$, $n\in {\mathbb N} - \{1\}$, of course is allowed to be different for different angles and triangles. Given a triangle $T$ with angles of this form, consider first the reflection group $\Gamma=\Gamma_T< Isom(H^2)$ generated  by reflections in the edges of $T$. Then $T$ is a fudnamental polygon for this group. The images of the triangle $T$ under $\Gamma$ tile the hyperbolic plane $H^2$ so that no $\Gamma$-wall in $H^2$ crosses $T$; this is an immediate consequence of being a fundamental domain. (Here, walls in $H^2$ are fixed lines of reflections which belong to $\Gamma$.) Now, suppose that $T'$ is another fundamental triangle of $\Gamma$. Then, after applying an element of $\Gamma$ to $T'$, we may assume that interiors of $T$ and $T'$ have nonempty intersection. But, since $T$ does not cross any walls, it follows that $T=T'$. 
Now, suppose that $\Gamma_{T}, \Gamma_{T'}$ are conjugate triangle reflection groups, $\Gamma_{T'}= g \Gamma_T g^{-1}, g\in Isom(H^2)$. Then $g^{-1}(T')$ is a fundamental triangle of $\Gamma$. By the above argument, it follows that $T, T'$ are congruent. 
Now, your groups $\Delta$ (under the angle assumption I made above) are index 2 subgroups in the triangle reflection groups. However, one can show that each $\Delta$ is an index 2 subgroup of a unique triangle reflection group. Thus, $\Delta$ is conjugate to $\Delta'$ iff the triangles are congruent. 
