How do you find the distance between $(2, π/3), (5, 2π/3)$ ?

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Answer 1 · 2017-06-11T15:54:40

The distance is $=4.63$

We convert the polar coordinates $(r,theta)$ into rectangular coordinates $(x,y)$

$(r,theta)$ , $=>$ , $(x,y)=(rcostheta, rsintheta)$

$(2,1/3pi)$ , $=>$ , $(2cos(1/3pi),2sin(1/3pi))=(1,sqrt3)$

and

$(5,2/3pi)$ , $=>$ , $(5cos(2/3pi),5sin(2/3pi))=(-5/2,5/2sqrt3)$

The distance is between $(1,sqrt3)$ and $(-5/2,5/2sqrt3)$

$d=sqrt((-5/2-1)^2+(5/2sqrt3-sqrt3)^2)$

$=sqrt(49/4+27/4)$

$=8.72/2$

$=4.36$

Answer 2 · 2017-06-11T16:05:04

Here is a graph of vectors from the origin to the two points:
Desmos.com

Please observe that the line connecting the two points forms a triangle:
Desmos.com

We can use the Law of Cosines to find the length of the blue line (side c):

$c^2 = a^2 + b^2 -2(a)(b)cos(theta)$

Where $a = 5, b = 2, and theta = (2pi)/3-pi/3 = pi/3$

$c^2 = 5^2 + 2^2 -2(5)(2)cos(pi/3)$

$c^2 = 25+4 - 2(10)(1/2)$

$c^2 = 19$

$c = sqrt19 larr$ answer

How do you find the distance between (2, π/3), (5, 2π/3)(2, π/3), (5, 2π/3)?