## Precalculus (6th Edition) Blitzer

Distance is, $2\sqrt{7}\ \text{units}$
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 $2\sqrt{7}\ \text{units}$.