I look at the following exercise of A. Pressley:
Show that similar spherical triangles are congruent.
I have no clue how to show it. Can you give an idea how I can do that?
In the book there is also this note:
In spherical geometry, similar triangles are congruent. This means that if two spherical triangles have vertices $a$, $b$, $c$ and $a'$, $b′$, $c′$, and if the angle of the first triangle at $a$ is equal to that of the second triangle at $a′$, and similarly for the other two angles, there is an isometry of $S^2$ that takes $a$ to $a′$, $b$ to $b′$ and $c$ to $c′$.
You want to show that the corresponding sides are equal. A tool that can be used for this purpose is spherical polar law of cosines which expresses each side of a spherical triangle as a function of angles. Thus, similar triangles have equal sides. (But apparently, your book allows you to assume that, according to the note you quoted.)
Having that, the isometry of $S^2$ can be described as follows: move one vertex $A$ to its counterpart $A'$, then rotate around $A'$ to bring $B$ to position $B'$ (possible since $|AB|=|A'B'|$), and then there are only two possible positions for $C$ and $C'$, symmetric about the arc $AB$. If needed, reflection across $AB$ brings $C$ to $C'$.