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)`