In $\triangle ABC$, $B = 45^\circ, C = 105^\circ, c=\sqrt{2}$. Find side $a$ and $b$.

$A = 180 - (45 + 105) = 30^\circ$ Using Law of Sines: $ \displaystyle \frac{a}{\sin A} = \frac{c}{\sin C} $ $ \displaystyle a = \frac{\sin 30^\circ}{\sin 105^\circ}\cdot \sqrt{2} $ Compute: $\sin 30^\circ = \frac12$ $\sin 105^\circ = \sin(60+45)=\frac{\sqrt{6}+\sqrt{2}}{4}$ $ \displaystyle a = \sqrt{2}\cdot \frac{1/2}{(\sqrt{6}+\sqrt{2})/4} = \frac{\sqrt{2}}{(\sqrt{6}+\sqrt{2})/2} = \frac{2\sqrt{2}}{\sqrt{6}+\sqrt{2}} $ After rationalizing: $ \displaystyle a=\sqrt{3}-1 $ Similarly, $ \displaystyle b=\sqrt{2}(\sqrt{3}-1) $