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Is it true that if a triangle on a unit sphere has 2 sides with equal length then their opposit angles must be equal? I think it is true. I think we can use the spherical sine law. Call the sides with equal lengths $a,b$ and their opposite angles $\alpha,\beta$. Then since $a=b$, $\sin\alpha=\sin\beta$. How do I then say for certain that $\alpha=\beta$? I know that the angles must be $\in (0,\pi)$ (right?). But how can I exclude the possibility of one angle being $\pi-$ the other angle?

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Say that you have $\triangle ABC$ with $AB = BC$. Now, observe that $\triangle CBA$ must be congruent to it. – Blue Feb 8 '12 at 14:36
@DayLateDon: I can see that, but How does it help to exclude the second case? – Karen Feb 8 '12 at 14:50
As the saying goes: "Corresponding Parts of Congruent Triangles are Congruent" ... so, note the corresponding parts of these two triangles. – Blue Feb 8 '12 at 14:53
@DayLateDon: Thanks :) Is this rigorous enough? I guess then i will have to prove the congruency. It is quite clear intuitively but how can I go about proving it? – Karen Feb 8 '12 at 15:17
Literally match the corresponding parts. I'm not just being cute with $\triangle CBA$, saying that the triangle is congruent to itself; I'm saying that the triangle is congruent to its flipped self, for reasons that should be Simple As ... um ... Spaghetti. (That said, my (mis?)understanding is that this oh-so-clever approach to a proof was only devised fairly recently by a proof-making computer, and had escaped notice of mathematicians going back to Euclid himself. So, it's okay to not quite get it at first; but I suspect there's an "ah ha!" moment coming.) – Blue Feb 8 '12 at 16:04

(bit late)... But since you were working on using the sine rule, you can say that $$\sin(\alpha )=\sin(\beta)\implies\alpha=\beta$$

because if,$$\alpha=\pi-\beta$$

The third angle is $0$, because the sum of the three is equal to $\pi$.

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