# What is the distance between (2 ,(5 pi)/6 ) and (-1 , (19 pi )/12 )?

Mar 18, 2018

$d = \sqrt{5 - 2 \setminus \sqrt{2}} \setminus \approx 1.47$

#### Explanation:

Distance for cartesian coordinate systems in two-dimensions would be given as:

$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

In polar coordinates, we have to change into using $r$ and $\setminus \theta$. We do this by using:

$x = r \cos \left(\setminus \theta\right)$
$y = r \sin \left(\setminus \theta\right)$

I'll spare the steps of deriving and give the result:

$d = \setminus \sqrt{{r}_{1}^{2} + {r}_{2}^{2} - 2 {r}_{1} {r}_{2} \cos \left(\setminus {\theta}_{1} - \setminus {\theta}_{2}\right)}$

We are given:
${r}_{1} = 2$, $\setminus {\theta}_{1} = \frac{5 \setminus \pi}{6}$
${r}_{2} = - 1$, $\setminus {\theta}_{2} = \frac{19 \setminus \pi}{12}$

The distance is then:

$d = \setminus \sqrt{{\left(2\right)}^{2} + {\left(- 1\right)}^{2} - 2 \left(2\right) \left(- 1\right) \cos \left(\frac{5 \setminus \pi}{6} - \frac{19 \setminus \pi}{12}\right)}$

$= \setminus \sqrt{4 + 1 + 4 \cos \left(\frac{- 9 \setminus \pi}{12}\right)}$

$= \setminus \sqrt{5 + 4 \cos \left(\frac{- 3 \setminus \pi}{4}\right)}$

$\implies \textcolor{g r e e n}{d = \sqrt{5 - 2 \setminus \sqrt{2}} \setminus \approx 1.47}$