In a $\triangle ABC$, if $a = \sqrt{3} - 1$, $b = \sqrt{3} + 1$ and $C = 60^\circ$, find the other side and the remaining two angles.
Given: In $\triangle ABC$, $a = \sqrt{3} - 1$, $b = \sqrt{3} + 1$, $C = 60^\circ$
By cosine rule,
$\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$
$\begin{aligned}
\implies c^2 & = a^2 + b^2 - 2 a b \cos C \\\\
& = \left(\sqrt{3} - 1\right)^2 + \left(\sqrt{3} + 1\right)^2 - 2 \left(\sqrt{3} - 1\right) \left(\sqrt{3} + 1\right) \cos \left(60^\circ\right) \\\\
& = 3 + 1 - 2 \sqrt{3} + 3 + 1 + 2 \sqrt{3} - 4 \times \dfrac{1}{2} \\\\
& = 6 \\\\
\implies c & = \sqrt{6}
\end{aligned}$
By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$
$\begin{aligned}
\implies \sin B & = \dfrac{b \sin C}{c} \\\\
& = \dfrac{\left(\sqrt{3} + 1\right) \sin \left(60^\circ\right)}{\sqrt{6}} \\\\
& = \dfrac{\left(\sqrt{3} + 1\right)}{\sqrt{6}} \times \dfrac{\sqrt{3}}{2} \\\\
& = \dfrac{\sqrt{3} + 1}{2 \sqrt{2}}
\end{aligned}$
$\implies$ $B = \sin^{-1} \left(\dfrac{\sqrt{3} + 1}{2 \sqrt{2}}\right) = 105^\circ$
In $\triangle ABC$, $A + B + C = 180^\circ$
$\therefore \;$ $A = 180^\circ - \left(B + C\right) = 180^\circ - \left(105^\circ + 60^\circ\right) = 15^\circ$