Points #(2 ,5 )# and #(3 ,4 )# are #( pi)/3 # radians apart on a circle. What is the shortest arc length between the points?

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Answer 1 · 2017-07-14T17:51:16

The length of the arc is #=1.48#

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The angle #theta=1/3pi#

The distance between the points is

#d=sqrt((3-2)^2+(4-5)^2)#

#=sqrt(1+1)#

#=sqrt2#

This is the length of the chord

So,

#d/2=rsin(theta/2)#

#r=d/(2sin(theta/2))#

#=sqrt2/(2sin(1/6pi))#

#=1.41#

The length of the arc is

#L=rtheta=1.41*1/3pi=1.48#

Answer 2 · 2017-07-14T18:03:10

#s = sqrt2pi/3#

Begin by finding the square of the length of the chord connecting the two points:

#c^2 = (3-2)^2+(4-5)^2#

#c^2 = 1^2+(-1)^2#

#c^2 = 2#

We can use a variant the Law of Cosines:

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

To find the radius of the circle, by letting, #c^2 = 2#, #a = b = r#, and #theta = pi/3#:

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

#2= 2r^2-2r^2(1/2)#

#2 = r^2#

#r=sqrt2#

We know that the arclength, s, between two points on a circle is the product of the radius and the radian measure of the central angle:

#s = rtheta#

Substitute the values for #r and theta#

#s = sqrt2pi/3#