If a spherical triangle has two equal sides then then it has two equal angles Suppose we have a spherical triangle $ABC$ in which two sides are equal, say $AC=BC$. I want to show that this necessarily means that the angles opposite these sides are equal. 

I've tried playing around with both the spherical cosine law and the spherical sine law, but neither of these seem to be getting me very far. 

Spherical cosine law: $$\sin(a)\sin(b)\cos(\gamma)=\cos(c)-\cos(a)\cos(b)$$ 

How is best to do this?
To prove such statement does not really require the fifth postulate, hence it is true both in Euclidean, spherical and hyperbolic geometry. Let $d$ be the geodetic distance on the sphere. $d(A,C)=d(B,C)$ implies that $A$ and $B$ belong to a geodetic circle centered at $C$. Let $M_C$ be the midpoint of the $AB$ arc on the sphere: by the definition of perpendicular bisector as a geometric locus, both $C$ and $M_C$ belong to the perpendicular bisector of $AB$ and the spherical triangles $AM_C C$, $BM_C C$ are congruent by SAS.
The figure is symmetrical about the angle bisector of the vertex angle.