Find the area of the triangle whose vertices are $\left(2, \dfrac{\pi}{3}\right)$, $\left(3, \dfrac{\pi}{2}\right)$, $\left(2, \dfrac{5 \pi}{6}\right)$
The given polar coordinates are $\;$ $A\left(r_1, \theta_1\right) = \left(2, \dfrac{\pi}{3}\right)$, $\;\;$ $B\left(r_2, \theta_2\right) = \left(3, \dfrac{\pi}{2}\right)$, $\;\;$ $C \left(r_3, \theta_3\right) = \left(2, \dfrac{5 \pi}{6}\right)$
Area of triangle $ABC$ is
$\begin{aligned}
= \Delta & = \dfrac{r_1 r_2 r_3}{2} \left\{\dfrac{\sin \left(\theta_3 - \theta_2\right)}{r_1} + \dfrac{\sin \left(\theta_1 - \theta_3\right)}{r_2} + \dfrac{\sin \left(\theta_2 - \theta_1\right)}{r_3} \right\} \\\\
& = \dfrac{2 \times 3 \times 2}{2} \left\{\dfrac{\sin \left(\dfrac{5 \pi}{6} - \dfrac{\pi}{2}\right)}{2} + \dfrac{\sin \left(\dfrac{\pi}{3} - \dfrac{5 \pi}{6}\right)}{3} + \dfrac{\sin \left(\dfrac{\pi}{2} - \dfrac{\pi}{3}\right)}{2} \right\} \\\\
& = 6 \times \left\{\dfrac{\sin \left(\dfrac{\pi}{3}\right)}{2} + \dfrac{\sin \left(\dfrac{-\pi}{2}\right)}{3} + \dfrac{\sin \left(\dfrac{\pi}{6}\right)}{2} \right\} \\\\
& = 6 \times \left\{\dfrac{\dfrac{\sqrt{3}}{2}}{2} - \dfrac{1}{3} + \dfrac{\dfrac{1}{2}}{2} \right\} \\\\
& = 6 \times \left\{\dfrac{\sqrt{3}}{4} - \dfrac{1}{3} + \dfrac{1}{4} \right\} \\\\
& = \dfrac{3 \sqrt{3} - 1}{2} \;\; \text{ square units}
\end{aligned}$