$2+$ |
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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 $C$.
The first method. We square both equalities and add them. We get $4=2+2(\sin A \cos B+\cos A \sin B)=2+2 \sin (A+B)$. Therefore,

$\sin (A+B)=1$, that is, $A+B=90^{\circ}$.

The 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, $C=90^{\circ}$.

The third method. Multiplying the original equalities, we get $\sin 2 A+\sin 2 B+2 \cos (A-B)=4$, from which $\sin 2 A=$ $\sin 2 B=\cos (A-B)=1$, that is,

$A=B=45^{\circ}$.

## Answer

$90^{\circ}$.