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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?

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    Have you tried solving the problem on a plane and then removing it and placing it on the sphere? – 2017-02-22
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    @Alephnull How can I do that? – 2017-02-22
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    The figure is symmetrical. – 2017-03-04

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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.

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The figure is symmetrical about the angle bisector of the vertex angle.

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    Far from being a proof, this should be a comment. – 2017-08-28