11.2. For the angles of triangle $ABC$, it is known that $\sin A + \cos B = \sqrt{2}$ and $\cos A + \sin B = \sqrt{2}$. Find the measure of angle $C$.
Answer: $90^{\circ}$.

First method. Squaring both equalities and adding them. After transformations, we get $\sin (A+B)=1$, that is, $A+B=90^{\circ}$.

Second method. Adding the original equalities, we get $\sin \left(A+45^{\circ}\right)+\sin \left(B+45^{\circ}\right)=2$, from which $\sin \left(A+45^{\circ}\right)=\sin \left(B+45^{\circ}\right)=1$. Therefore, $A=B=45^{\circ}$. Hence, $C=90^{\circ}$.