Question

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

Answer 1

`d = sqrt{5-2\sqrt(2)} \approx 1.47`

Explanation:

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

`d = sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}`

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

`x = rcos(\theta)`
`y = rsin(\theta)`

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

`d = \sqrt{r_{1}^{2} + r_{2}^{2} - 2r_{1}r_{2} cos (\theta_{1} -\theta_{2})}`

We are given:
`r_1 = 2`, `\theta_1 = (5\pi)/6`
`r_2 = -1`, `\theta_2 = (19\pi)/12`

The distance is then:

`d = \sqrt{(2)^2+(-1)^2-2(2)(-1) cos((5\pi)/6-(19\pi)/12)}`

`= \sqrt{4+1+4cos((-9\pi)/12)}`

`= \sqrt{5+4 cos((-3\pi)/(4))}`

`=> color(green)(d = sqrt{5-2\sqrt(2)} \approx 1.47)`