Given a triangle $A B C$, where $A C=\sqrt{2}, B C=1, \angle B=45^{\circ}$. Find the angle $A$.
Use the Law of Sines.

## Solution

By the Law of Sines, $\frac{1}{\sin \angle A}=\frac{\sqrt{2}}{\sin 45^{\circ}}$, from which $\sin \angle A=1 / 2$. Therefore, $\angle A=30^{\circ}$ or $\angle A=150^{\circ}$. The second

case contradicts the theorem of the sum of the angles of a triangle. Therefore, $\angle A=30^{\circ}$.

## Answer

$30^{\circ}$.