Points #(6 ,2 )# and #(1 ,5 )# are #(2 pi)/3 # radians apart on a circle. What is the shortest arc length between the points?

The required in the problem is to find the shortest arc length #s# subtended by the central angle #theta=(2pi)/3#

The chord #l# which has length of the distance between the two points (1, 5) and (6, 2)

#l=sqrt34#

This chord forms the triangle with sides #l#, #r#, #r# and with angle #theta=(2pi)/3# opposite side #l#.

By the cosine law, we have the equation

#l=sqrt((r^2+r^2-2*r*r*cos theta))#
#sqrt34=sqrt((r^2+r^2-2*r*r*cos ((2pi)/3))#
#sqrt34=sqrt((2r^2-2r^2(-1/2)))=sqrt(3*r^2)#

Solving for #r#

#r=sqrt(34)/sqrt3=sqrt(102)/3#

Solve for the arc length #s#

#s=r*theta#
#s=sqrt(102)/3*(2pi)/3#
#s=(2sqrt(102)pi)/9=7.05078" "#units

God bless....I hope the explanation is useful.

Given angle= #theta=(2pi)/3#;
Two points on the circle #A(1,5)# & #B(6,2)#
Find the Arc Length?
Definition and Principles: Arc Length #L=thetar=(2pi)/3r#
Need to find #r#?

Let's use the geometric figure, we see that since #hat(AB)=(2pi)/3# it follows that #/_ABC=(2pi)/3 or 120^o#
Using the distance formula
#bar(AB) = sqrt((1-6)^2+(5-2 )^2)=sqrt(5^2+3^2)=sqrt(34)#
Now ABC is an isosceles triangle so that altitude of the triangle bisect the #bar(AB)= bar(AH)+bar(HB)=sqrt(34)/2#

Now given we have an isosceles triangle #ABC_Delta# the altitude bisects the triangle into two right angle traingles #ABH and BHC# with angles (30, 60, 90), with triangle sides ratio of #1:sqrt(3):2#
Thus #bar(AH):sqrt(3)=bar(AB):2; 1/2sqrt(34):sqrt(3)=AB:2#
#bar(AB)=sqrt(34)/sqrt(3)= sqrt(102)/3#; #AB=r=sqrt(102)/3#
Now with radius known use the Arc Length formula to calculate Arc Length.
#L=(2pi)/3sqrt(102)/3= (2pi)/9sqrt(102)#