Find the area of $\triangle ABC$ when $a = \sqrt{3}$, $b = \sqrt{2}$ and $c = \dfrac{\sqrt{6} + \sqrt{2}}{2}$
Given: $\;$ $a = \sqrt{3}$, $\;$ $b = \sqrt{2}$, $\;$ $c = \dfrac{\sqrt{6} + \sqrt{2}}{2}$
By cosine rule,
$\begin{aligned}
\cos C & = \dfrac{a^2 + b^2 - c^2}{2 a b} \\\\
& = \dfrac{3 + 2 - \left(\dfrac{6 + 2 + 4 \sqrt{3}}{4}\right)}{2 \times \sqrt{3 \sqrt{2}}} \\\\
& = \dfrac{5 - \left(2 + \sqrt{3}\right)}{2 \sqrt{6}} \\\\
& = \dfrac{3 - \sqrt{3}}{2 \sqrt{6}} \\\\
& = \dfrac{\sqrt{3} - 1}{2 \sqrt{2}} \;\;\; \left[\text{dividing numerator and denominator by } \sqrt{3}\right]
\end{aligned}$
$\implies$ $C = \cos^{-1} \left(\dfrac{\sqrt{3} - 1}{2 \sqrt{2}}\right) = 75^\circ$
Now, $\;$ area of $\triangle ABC$ $= \Delta = \dfrac{1}{2}a b \sin C$
$\begin{aligned}
i.e. \; \Delta & = \dfrac{1}{2} \times \sqrt{3} \times \sqrt{2} \times \sin \left(75^\circ\right) \\\\
& = \dfrac{1}{2} \times \sqrt{6} \times \dfrac{\left(\sqrt{3} + 1\right)}{2 \sqrt{2}} \\\\
& = \dfrac{\sqrt{3} \left(\sqrt{3} + 1\right)}{4} \\\\
& = 1.183 \; \text{sq units}
\end{aligned}$