A square with side a is inscribed in a sector with a central angle of 120 degrees. Find the radius of the sector.

The AOD triangle is isosceles, then the AOD angle = ADO = (180 – AOD) / 2 = (180 – 120) / 2 = 30.

Then the angle BAO = BAD + DAO = 90 + 30 = 120.

In an isosceles triangle, AOD, OH is the height, median and bisector, then AH = DH = a / 2 cm. OA = (a / 2) / √3 / 2 = a / √3 cm.

In triangle ABO, side AB = a cm, OA = a / √3, OB = R.

Then, by the cosine theorem: R ^ 2 = AB ^ 2 + AO ^ 2 – 2 * AB * AO * Cos120 = a ^ 2 + (a ^ 2/3) – 2 * a * (a / √3) * ( -1/2) = a ^ 2 + a ^ 2/3 + (a ^ 2 / √3) = a ^ 2 * (1 + (1/3) + (1 / √3)) = a ^ 2 * ((3 * √3 + √3 + 3) / 3 * √3) = a ^ 2 * (3 + 1 + √3) / 3 = a ^ 2 * (4 + √3) / 3.

R = a * √ (4 + √3) / √3.

Answer: The radius of the sector is a * √ (4 + √3) / √3.