The sides of a triangle are $\sin \alpha ,\cos \alpha$ and $\sqrt{1+\sin \alpha \cos \alpha }$ for some $0<\alpha <\frac{\pi }{2}.$ Then the greatest angle of the triangle is

Let the sides be $a=\sin \alpha ,b=\cos \alpha ,$ and $c=\sqrt{1+\sin \alpha \cos \alpha }$.
Note that $c^2=1+\sin \alpha \cos \alpha .$
Also, $a^2+b^2={\sin }^2\alpha +{\cos }^2\alpha =1.$
Since $\alpha \in (0,\pi /2),$ both $\sin \alpha$ and $\cos \alpha$ are positive, making $\sin \alpha \cos \alpha >0.$
Thus $c^2>a^2+b^2,$ which means $c$ is the longest side and the angle $C$ opposite to it is the greatest angle.
Using the Law of Cosines:
$\cos C=\frac{a^2+b^2-c^2}{2ab}$.
$\cos C=\frac{1-(1+\sin \alpha \cos \alpha )}{2\sin \alpha \cos \alpha }=\frac{-\sin \alpha \cos \alpha }{2\sin \alpha \cos \alpha }=-\frac{1}{2}$.
$C=120^{\circ }$.