[tex]\displaystyle\\
\text{Aplicam teorema sinusurilor.}\\\\
\frac{BC}{\sin A} = \frac{AB}{\sin C} \\\\
\frac{2}{\sin 45^o} = \frac{ \sqrt{2} }{\sin C} \\\\
\sin C = \frac{\sqrt{2}\times \sin 45^o}{2} = \frac{\sqrt{2}\times \dfrac{\sqrt{2}}{2} }{2} =\frac{\dfrac{\sqrt{2}\times\sqrt{2}}{2} }{2} =\frac{~\dfrac{2}{2} ~}{2} = \boxed{\frac{1}{2} }\\\\
\Longrightarrow~~m(\sphericalangle C) = \arcsin \left(\frac{1}{2} \right) = \boxed{\bf 30^o}[/tex]
A doua solutie (180° - 30° = 150°) nu este admisa avand deja un unghi de 45°.
150° + 45° = 195° mai mare decat 180° (suma unghiurilor dintr-un triunghi).
