Points #(8 ,5 )# and #(7 ,6 )# are #( pi)/3 # radians apart on a circle. What is the shortest arc length between the points?

1 Answer
Douglas K.
Nov 19, 2016

#s = sqrt(2)pi/3#

Explanation:

The length of the chord is:

#c = sqrt((7 - 8)^2 + (6 - 5)^2)#

#c = sqrt(2)#

Two radii, each drawn from the center to its respective end of the chord, form a triangle with the chord, therefore, we can use a variant of the Law of Cosines where #a = b = r#:

#c^2 = r^2 + r^2 - 2(r)(r)cos(theta)#

Substitute #c = sqrt(2) and theta = pi/3#:

NOTE: At this point, we should realize that an isosceles triangle with the third angle equal to #pi/3# is an equilateral triangle but, let's proceed as an example of how to solve the problem, when it is not this special case:

#(sqrt(2))^2 = 2r^2 - 2r^2cos(pi/3)#

#r^2 = (sqrt(2))^2/(2(1 - cos(pi/3))#

#r^2 = 2#

#r = sqrt(2)#

The arc length is, #s = rtheta#

#s = sqrt(2)pi/3#

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