If the sides of a triangle are $\sin a, \cos a$,$\sqrt{1+\sin \alpha \cos \alpha }$,$0$$<\alpha <\frac{\pi }{2}$ the largest angle is

 

Solution:

The largest side is $c=\sqrt{1+\sin \alpha \cos \alpha }$

$\cos C=\frac{a^2+b^2-c^2}{2ab}$

$=\frac{{\sin }^2\alpha +{\cos }^2\alpha -(1+\sin \alpha \cos \alpha )}{2\sin \alpha \cos \alpha }=-\frac{1}{2}$

$\therefore C=120°$.