What is the distance between the following polar coordinates?: # (8,(-21pi)/12), (5,(-3pi)/8) #

Question

What is the distance between the following polar coordinates?: #   (8,(-21pi)/12), (5,(-3pi)/8)   #

William Abdalla · Accepted Answer

#10.937#

First #(8,-(21pi)/12)# can be simplified a bit #(8, -(7pi)/4)#, and since #-(7pi)/4# is coterminal to #pi/4#, we'll use #(8,pi/4)# as an equivalent point. The other point we'll keep as #(5,-(3pi)/8)#.

If we plot the points and use them as two vertices of a triangle with the origin as the third vertex, we have sides 8 and 5 with an angle of #pi/4 + (3pi)/8 = (5pi)/8# between them.

For this triangle we can use the Law of Cosines to find the side opposite #(5pi)/8#, which is the distance between the given points:

#c=sqrt(a^2+b^2-2abcos(C))#

#c=sqrt(8^2+5^2-2(8)(5)cos((5pi)/8))#

#c approx 10.937#

This video might also be helpful.

Christopher Allers · Answer

undefined To find the distance between two polar coordinates, you can use the formula:

$d = \sqrt{r_1^2 + r_2^2 - 2r_1r_2\cos(\theta_2 - \theta_1)}$

Given the polar coordinates:
$r_1 = 8$, $\theta_1 = -\frac{21\pi}{12}$
$r_2 = 5$, $\theta_2 = -\frac{3\pi}{8}$

Plugging these values into the formula:

$d = \sqrt{8^2 + 5^2 - 2(8)(5)\cos\left(-\frac{3\pi}{8} - \left(-\frac{21\pi}{12}\right)\right)}$

$d = \sqrt{64 + 25 - 80\cos\left(-\frac{3\pi}{8} + \frac{7\pi}{4}\right)}$

$d = \sqrt{89 - 80\cos\left(\frac{\pi}{8}\right)}$

$d = \sqrt{89 - 80\left(\frac{\sqrt{2 + \sqrt{2}}}{2}\right)}$

$d \approx \sqrt{89 - 40\sqrt{2 + \sqrt{2}}}$

Therefore, the distance between the two polar coordinates is approximately $d \approx 4.019$.