The given points are $i, 3i, 1 + 2i$
I know that the distance for points on a vertical line can be found by using the formula
$$\ln\left|\frac{y_2}{y_1}\right|$$
So the distance between points $ i$ and $3i$ would be $\ln 3$. How can I get the length between the other sides and the angles of the triangle?
The two flavors of unit speed geodesics, in the upper half plane, are the one you know, $$ \alpha(t) = A + i e^t, $$ for real constant $A$, and $$ \beta(t) = A + B \tanh t + i B \operatorname{sech} t $$ for real constants $A,B$ with $B > 0.$
Since people do not appear to be familiar with the second type, I request that you ask your instructor about it.
My impression from MSE questions from the past week or two is that people are being taught that Möbius transformations (with positive matrix determinant and real entries) give isometries. If you apply the map $-1/z$ to the vertical line, $ \alpha(t) = A + i e^t, $ you get one example of the second type, which travels along a semicircle with center on the real line. This can be moved about and $B$ changed with additional Möbius transformations.
You should be able to finish from here. You need to find the $A,B$ and two $t$ values along two circular arcs. The difference in $t$ values is the length of that arc in the model.
Or, you can use a Möbius transformation that maps two points to other points with the same real part, in which case you can use your little logarithm. Probably about the same amount of work.