If $\;$ $a = 2$, $c = \sqrt{3} + 1$, and $A = 45^\circ$, solve the triangle.
Given: $\;$ $a = 2$, $\;$ $c = \sqrt{3} + 1$, $\;$ $A = 45^\circ$
Now, $\;$ $c \sin A = \left(\sqrt{3} + 1\right) \sin 45^\circ = \dfrac{\sqrt{3} + 1}{\sqrt{2}} = 1.932 < a$
$\implies$ There are two values of $C$, $0^\circ < C < 90^\circ$ and $90^\circ < C < 180^\circ$
In $\triangle ABC$, by sine rule, $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$
$\implies$ $\dfrac{2}{\sin 45^\circ} = \dfrac{\sqrt{3} + 1}{\sin C}$
i.e. $\;$ $\sin C = \dfrac{\left(\sqrt{3} + 1\right) \sin 45^\circ}{2} = \dfrac{\sqrt{3} + 1}{2 \sqrt{2}}$
$\implies$ $C = 75^\circ$ $\;$ or $\;$ $105^\circ$
In $\triangle ABC$, $\;$ $A + B + C = 180^\circ$ $\implies$ $B = 180^\circ - \left(A + C\right)$
When $\;$ $C = 75^\circ$, $\;$ $B = 180^\circ - \left(45^\circ + 75^\circ\right) = 60^\circ$
When $\;$ $C = 105^\circ$, $\;$ $B = 180^\circ - \left(45^\circ + 10^\circ\right) = 30^\circ$
Also, by sine rule,
$b = \dfrac{a \sin B}{\sin A}$
When $\;$ $B = 60^\circ$, $\;$ $b = \dfrac{2 \times \sin 60^\circ}{\sin 45^\circ} = \dfrac{2 \times \dfrac{\sqrt{3}}{2}}{\dfrac{1}{\sqrt{2}}} = \sqrt{6}$
When $\;$ $B = 30^\circ$, $\;$ $b = \dfrac{2 \times \sin 30^\circ}{\sin 45^\circ} = \dfrac{2 \times \dfrac{1}{2}}{\dfrac{1}{\sqrt{2}}} = \sqrt{2}$
$\therefore \;$ $B = 60^\circ$, $\;$ $C = 75^\circ$, $\;$ $b = \sqrt{6}$
or, $\;$ $B = 30^\circ$, $\;$ $C = 105^\circ$, $\;$ $b = \sqrt{2}$