Answer
Distance is, $2\sqrt{7}\ \text{units}$
Work Step by Step
The distance between two points in rectangular coordinates $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is given by
$d=\sqrt{{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}}$
Substitute, ${{r}_{1}}=2,{{\theta }_{1}}=\frac{5\pi }{6}$ and ${{r}_{2}}=4,{{\theta }_{2}}=\frac{\pi }{6}$ in the formula as,
$\begin{align}
& d=\sqrt{{{2}^{2}}+{{4}^{2}}-2\times 2\times 4\times \cos \left( \frac{5\pi }{6}-\frac{\pi }{6} \right)} \\
& =\sqrt{4+16-16\cos \left( \frac{2\pi }{3} \right)} \\
& =\sqrt{20-16\left( -\frac{1}{2} \right)} \\
& =\sqrt{28}
\end{align}$
This implies:
$d=2\sqrt{7}\text{units}$
Therefore, the distance between $\left( 2,\frac{5\pi }{6} \right)$ and $\left( 4,\frac{\pi }{6} \right)$ is $2\sqrt{7}\ \text{units}$.
