# What is the distance between the following polar coordinates?:  (4,(-8pi)/3), (-5,(11pi)/6)

Sep 28, 2016

$d = \sqrt{41}$

#### Explanation:

Convert $\left(4 , \frac{- 8 \pi}{3}\right)$ to rectangular coordinates

$\left(4 \cos \left(\frac{- 8 \pi}{3}\right) , 4 \sin \left(\frac{- 8 \pi}{3}\right)\right)$

Convert $\left(- 5 , \frac{11 \pi}{6}\right)$ to rectangular coordinates

$\left(- 5 \cos \left(\frac{11 \pi}{6}\right) , - 5 \sin \left(\frac{11 \pi}{6}\right)\right)$

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

d = sqrt((4cos((-8pi)/3) + 5cos((11pi)/6))^2 + (4sin((-8pi)/3) + 5sin((11pi)/6))^2

$d = \sqrt{41}$

Sep 28, 2016

$\text{The Dist.=} \sqrt{41} \approx 6.40$

#### Explanation:

The Distance $A B$ btwn. the polar pts. $A \left({r}_{1} , {\theta}_{1}\right)$ and

$B \left({r}_{2} , {\theta}_{2}\right)$ is,

AB=sqrt(r_1^2+r_2^2-2r_1r_2cos(theta_1-theta_2).

Accordingly, the reqd. dist.

=sqrt{4^2+(-5)^2-2(4)(-5)cos(11pi/6+8pi/3)

=sqrt(16+25+40cos(27pi/6)

=sqrt(41+40cos(9pi/2)

$= \sqrt{41 + 0}$

$= \sqrt{41.}$

$\approx 6.40$.