I'm having trouble with the following question from spherical geometry:
My book sets R=1.
Let $\alpha$, $\beta$, $\gamma$ be the side lengths of a spherical triangle $ \triangle PQR$ and $a, b, c$ the opposite angles. Use the main formula:
$\cos \alpha = \cos \beta \cos \gamma + \sin \beta \sin \gamma \cos a$
to prove that $|\beta-\gamma|<\alpha <\beta + \gamma $, and that $ \alpha +\beta +\gamma < 2 \pi $
The second inequality follows by contradiction. Assume that $ \alpha +\beta +\gamma = 2 \pi $, then using the angle sum formula for the cosine we find that:
$\cos \alpha = \cos \beta \cos \gamma - \sin \beta \sin \gamma$
and that $cos(a)=-1$, this only happens for angle $\pi$. Which would contradict the fact that PQR is a triangle. I notice that $\alpha <\beta + \gamma $ follows from the triangle inequality for metric spaces, the other side follows from inverse triangle inequality. $|\beta-\gamma|<\alpha $, but the last step is:
Prove that three sides $\alpha, \beta, \gamma <\pi$, satisfying these two inequalities are sides of a spherical triangle. Any pointers?