Question 1210153
<pre>

{{{drawing(320,400,-4,4,-1,9, 

triangle(0,0,-3,3sqrt(3),3,3sqrt(3)),

locate(0,5.2,2), 
locate(1.24,3,2), locate(-1.5,3,2),
locate(-.2,1.2,60^o),
arc(0,0,12,-12,60,120),

locate(-3.2,5.5,A),
locate(3.1,5.5,B)


  )}}}

The triangle is equilateral, so the central angle is 60<sup>o</sup>,
so the arc is {{{60^o/360^o = 1/6}}} of the circumference of a great circle.

The circumference of a great circle is {{{2*pi*r=2*pi*2=4*pi}}}

The length of the arc is {{{expr(1/6)*4pi}}} or {{{expr(2/3)pi}}}.

Edwin</pre>