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

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Answer 1 · 2016-09-28T19:00:45

$d = sqrt(41)$

Convert $(4, (-8pi)/3)$ to rectangular coordinates

$(4cos((-8pi)/3), 4sin((-8pi)/3))$

Convert $(-5, (11pi)/6)$ to rectangular coordinates

$(-5cos((11pi)/6), -5sin((11pi)/6))$

$d = sqrt((x_1 - x_2)^2 + (y_1 - y_2)^2)$

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

$d = sqrt(41)$

Answer 2 · 2016-09-28T19:33:27

$"The Dist.="sqrt41~~6.40$

The Distance $AB$ btwn. the polar pts. $A(r_1,theta_1)$ and

$B(r_2,theta_2)$ is,

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

Accordingly, the reqd. dist.

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

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

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

$=sqrt(41+0)$

$=sqrt41.$

$~~6.40$ .

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