I came across this problem in some (maybe) high school book:
Let $ABC$ be an isosceles triangle s.t. $AB=AC$. Also, $\alpha>\beta$. It is known/given:
Find the base angles of $\triangle ABC$.
I've tried pretty much everything involving the law of cosines/sines, and also auxiliary constructions of the normal to $BC$ (in $\triangle ABC$), which enables looking at the circumscribed circles of the two halves of $\triangle ABC$.
I will be glad to hear any insight about this problem. I think I'm missing something very elementary, as I didn't find the second equation too helpful. (The first one is obviously true for all isosceles triangles.)