Let $a = \sin \alpha ,b = \cos \alpha$ and $c = \sqrt {1 + \sin \alpha \cos \alpha }$ Clearly $a$ and $b < 1$ but $c > 1$ as $\,\,\,\sin \alpha > 0$ and $\cos \alpha > 0$ $\therefore$ $c$ is the greatest side and greatest angle is $C$ $\therefore$ $\cos C = {{{a^2} + {b^2} - {c^2}} \over {2ab}}$ $= {{{{\sin }^2}\alpha + {{\cos }^2}\alpha - 1 - \sin \alpha \cos \alpha } \over {2\,\sin \alpha \cos \alpha }}$ $= - {1 \over 2}$ $\therefore$ $C = {120^ \circ }$