I will denote $\alpha ,\beta, \gamma$ by $a,b,c$ .W.L.O.G $a\geq b\geq c$. Since $b+c>a$, we have $a< \frac{\pi}{2}$.For existence of a triangle with side lengths as $\sin a, \sin b , \sin c$ its necessary and sufficient to have $\sin b + \sin c > \sin a$. If $b+c<\frac{\pi}{2}$, then $\sin b + \sin c >\sin(b+c) > \sin a$, If $b+c>\frac{\pi}{2}$, then $a< \pi - b-c <\frac{\pi}{2}$ so again $\sin b + \sin c >\sin(b+c) > \sin a$.

My proof of the first part is the same as Learner 94,and it may essentially the same for all.However can someone give a proof of the second part.Or,I think I am missing something,maybe its too easy.

why don't we dicuss it i think it is very interesting