Prove that isometries of $S^2$ preserve internal angles of spherical triangles Let $ABC$ be a triangle on $S^2$ with internal angles $\alpha, \beta, \gamma$ and opposite side lengths $a, b, c$ respectively. Let $T$ be an isometry of $S^2$
Using the spherical cosine rule, we see that $\cos \alpha = \frac{\cos a - \cos b \cos c}{\sin b \sin c}$ is preserved, which means that if $\alpha'$ is the image of  $\alpha$ under $T$, then $\alpha' = \alpha$ or $\alpha' = 2\pi - \alpha$. I'm struggling to prove that the former is true and not the latter. In other words, I'm struggling to prove that an isometry of $S^2$ doesn't "flip" the triangle inside out so that it looks the same, but its inside has swapped with its outside.
I'm allowed to use spherical trigonometry arguments, coordinate-geometric arguments, synthetic-geometric arguments, as well as e.g. facts about the distance metric on the sphere surface. However the only fact about isometries that I can use is that they preserve lengths.
 A: Updated argument disallowing Riemannian geometry:
The issue in this question seems to me to be to be expressible in "synthetic" geometry terms as follows: Given a point $A$ and two short line segments $\gamma,\delta$ intersecting only at a common endpoint coinciding with $A$, and assuming that $\gamma,\delta$ do not form a straight angle at $A$, how do you distinguish the "inner" angle $\angle \gamma\delta$ from the "outer" or "reflex" angle?
Here's a rough idea. By extending $\gamma,\delta$, we may assume that they are each great half-circles, having one common endpoint at $A$ and opposite common endpoint at $A'$ which is antipodal to $A$. So $\gamma \cup \delta$ subdivides $S^2$ into two pieces which I'll call "lunes". Neither of these two lunes is a hemisphere, because we assume that $\gamma,\delta$ do not form a straight angle at $A$. We wish to distinguish one as the "inner lune" and the other as the "outer" or "reflex lune". 
I think the idea is exactly one of the two lunes is "convex", and we take one that to be the inner lune. Here we should define a subset $X$ of the sphere to be "convex" if for any pair of non-antipodal points $x,y \in X$, the unique shortest geodesic segment $\overline{xy}$ is also contained in $X$. Convexity is clearly an invariant under isometry. 
Let me add that the same method should work in Euclidean geometry, where $\gamma,\delta$ are rays with a common endpoint that do not form a straight angle: the union $\gamma \cup \delta$ subdivides the plane into two pieces, one of which is convex and the other is not.
Older argument allowing Riemannian geometry:
Isometries in Riemannian geometry preserve area. The total area of $S^2$ is $4\pi$, the area of the "inside" of the triangle is $<2\pi$, and the area of the outside is $> 2 \pi$. Therefore the isometry must take the inside of the $ABC$ triangle to the inside of the $A'B'C'$ triangle ($A'=T(A)$ etc.).
You can apply this same area argument to a "bigon" (or "lune" in my updated argument above) consisting of a pair of half-great-circles meeting a pair of antipodal points, with the conclusion (expressed colloquially) that an isometry takes the "inside" of an angle to the "inside" of the image angle.
