Square in a Circular Sector

Assume the sector to be a part of the circle: x²+y²=100². Let (p.q) be a point on the right hand side straight line, y= $\frac{x}{√3}$ . Or q(√3)=p. Further let x be the side of the square; then x=2p & (p,q+x) lies on our circle. Hence, p²+(q+x)²=100². On substitution one obtains, x= $\frac{100√3}{√(4+√3)}$ = 72.3445

Let the side length of the square be 2a. Let A and B be the points where the top-right and the bottom-right corner of the square intersect the sector, O be the origin, C be the point where the Y-axis intersects the square initially. Extend AB to meet the X-axis at D.

$\angle CBO$ = $90$ - $\frac{120}{2}$ = $30$

Since the square is placed symmetrically, $CB$ = $\frac{2a}{2}$ = $a$ .

$OC$ = $BD$ = $a\times tan30$ = $\frac{a}{\sqrt3}$ .

$AD$ = $AB$ + $BD$ = $2a$ + $\frac{a}{\sqrt3}$ .

$OD$ = $CB$ = $a$ .

$AO$ =radius of the sector= $100$ .

By the Pythagorean Theorem,

$AD^{2}$ + $OD^{2}$ = $AO^{2}$

$(2a+\frac{a}{\sqrt3})^{2}$ + $a^{2}$ = $10000$

Solving we get, $2a$ $\approx$ $72.3445$