$3$ square

We see isosceles right triangle $BFM$ with an angle of $45^{\circ}$ at the base. And $\angle \alpha =\angle MBN$ , $\angle \beta =\angle FBN$ and $\angle FBN+\angle MBN=45^{\circ}$ .

$\large \sin { (\alpha +\beta ) } =\sin { \alpha } \cos { \beta } +\cos { \alpha } \sin { \beta } \\ \large \sin { (\alpha +\beta ) } =\frac { a }{ a\sqrt { 10 } } \cdot \frac { 2a }{ a\sqrt { 5 } } +\frac { 3a }{ a\sqrt { 10 } } \cdot \frac { a }{ a\sqrt { 5 } } \\ \large \sin { (\alpha +\beta ) } =\frac { 2 }{ \sqrt { 50 } } +\frac { 3 }{ \sqrt { 50 } } \\ \large \sin { (\alpha +\beta ) } =\frac { 5 }{ \sqrt { 50 } } \\ \large \sin { (\alpha +\beta ) } =\frac { 5 }{ 5\sqrt { 2 } } \\ \sin { (\alpha +\beta ) } =\frac { 1 }{ \sqrt { 2 } } \\ \sin { (\alpha +\beta ) } =\frac { 1 }{ 2 } \sqrt { 2 } \\ \large \therefore \alpha +\beta ={ 45 }^{ \circ }$