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

2 Answers
Jun 11, 2017

The distance is =4.63

Explanation:

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

Jun 11, 2017

Here is a graph of vectors from the origin to the two points:
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Explanation:

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