This is a modified version of this question, and as such I'm using similar wording and visuals. Given:
- $α_2$, one of the two angles which the vertex $A$ is split by the median $m$;
- $\overline{AD}$, the length of the segment of a triangle external to $ABC$;
- $\overline{BM}$, the length of half of the side split by the median;
- $θ$, the angle opposite to the shared side $\overline{AB}$;
Find the value of the angles $γ$ and $b_2$.
Similarly to the original question, the proportion $\frac{\sin{α_1}}{\sin{α_2}}=\frac{\sin{β_1}}{\sin{γ}}$ holds true. And it's still true that $β_1 = 180 - α_1 - α_2 - γ$. However, unlike the original scenario, we don't know the value of $α_1$.
With the external angle theorem, we know that $α_1 = θ + β_2 - α_2$, but now we have the $β_2$ variable to resolve.
I figured we could use the law of sines to establish $\frac{\sin{β_2}}{\overline{AD}}=\frac{\sin{θ}}{\overline{AB}}$, but after a few variable swaps the problem ultimately seems to loop back to $α_1$.
I feel that this problem is solvable, as the given parameters uniquely identify the two triangles. But I can't find out how to take it further than I have.