If $a = 2$, $b = 1 + \sqrt{3}$ and $C = 60^\circ$, solve the triangle.
Given: $\;$ $a = 2$, $\;$ $b = 1 + \sqrt{3}$, $\;$ $C = 60^\circ$
In $\triangle ABC$, by cosine rule,
$\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$
$\begin{aligned}
\implies c^2 & = a^2 + b^2 - 2 a b \cos C \\\\
& = \left(2\right)^2 + \left(1 + \sqrt{3}\right)^2 - 2 \times 2 \times \left(1 + \sqrt{3}\right) \cos 60^\circ \\\\
& = 4 + 1 + 3 + 2 \sqrt{3} - 2 - 2 \sqrt{3}
\end{aligned}$
i.e. $\;$ $c^2 = 6$ $\implies$ $c = \sqrt{6}$
By sine rule,
$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$
$\therefore \;$ We have, $\;$ $\dfrac{2}{\sin A} = \dfrac{\sqrt{6}}{\sin 60^\circ}$
i.e. $\;$ $\sin A = \dfrac{2 \sin 60^\circ}{\sqrt{6}} = 2 \times \dfrac{\sqrt{3}}{2} \times \dfrac{1}{\sqrt{6}} = \dfrac{1}{\sqrt{2}}$
$\implies$ $A = \sin^{-1} \left(\dfrac{1}{\sqrt{2}}\right) = 45^\circ$
and $\;$ $\dfrac{1 + \sqrt{3}}{\sin B} = \dfrac{\sqrt{6}}{\sin 60^\circ}$
i.e. $\;$ $\sin B = \dfrac{\left(1 + \sqrt{3}\right) \sin 60^\circ}{\sqrt{6}} = \dfrac{\left(1 + \sqrt{3}\right) \sqrt{3}}{2 \sqrt{6}} = \dfrac{1 + \sqrt{3}}{2 \sqrt{2}}$
$\implies$ $B = \sin^{-1} \left(\dfrac{1 + \sqrt{3}}{2 \sqrt{2}}\right) = 75^\circ$
$\therefore \;$ $c = \sqrt{6}$, $\;$ $A = 45^\circ$, $\;$ $B = 75^\circ$