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

Solution. Square both equalities and add them. Then use the fundamental trigonometric identity, and after transformations, we get $2 \sin \angle A \cos \angle \mathrm{B}+2 \sin \angle B \cos \angle \mathrm{A}=2 \sin (\angle A+\angle \mathrm{B})=2 \quad, \quad$ that is, $\quad \sin (\angle A+\angle \mathrm{B})=1$. Therefore, $\angle A+\angle \mathrm{B}=90^{\circ}, \angle \mathrm{C}=180^{\circ}-(\angle A+\angle \mathrm{B})=90^{\circ}$.