Question 1085256
Here is a sketch, with the midpoint of OA labeled as R:
{{{drawing(300,200,-1.05,1.05,-.2,1.2,
circle(0,0,0.015),circle(0,0,1),
red(triangle(0.866,0.5,-0.866,0.5)),
red(triangle(0,0.5,0,0,0,1)),
red(rectangle(0,0.5,0.05,0.45)),
locate(-0.02,-0.02,O),locate(-0.02,1.1,A),
locate(-0.91,0.58,P),locate(0.88,0.58,Q),
locate(0.02,0.6,R),circle(0,0.5,0.015)
)}}} Take a good look at right triangle OQR.
The lengths {{{OP=OA=OQ=R}}} are the radius, {{{R}}} , of the circle.
{{{OR=RA=(1/2)OA=(1/2)R=(1/2)OQ}}} is half of that radius.
That tells you that the short leg of right triangle OQR is half the hypotenuse.
It means that
{{{RQ=sqrt(3)R/2}}} (from applying the Pythagorean theorem).
It also means (considering trigonometric ratios for the angles of OQR) that
OQR is a 30-60-90 triangle, with a {{{60^o}}} angle {{{ROQ=AOQ}}} at O,
which makes angle {{{POQ}}} and arc {{{PAQ}}} measure {{{2*60^o=120^o}}} ,
or {{{1/3}}} of the whole circle.
If {{{2pi/3}}} is the length of arc {{{PAQ}}} ,
then {{{3(2pi/3)=2pi}}} is {{{2pi*R}}} , the length of the circumference.
Then, {{{R=1}}} , {{{RQ=sqrt(3)*1/2=sqrt(3)/2}}} , and {{{PQ=2RQ=2(sqrt(3)/2)=highlight(sqrt(3)=about1.732)}}} .