A regular triangle ABC is inscribed in a circle with center O, OB = √3 / 6. Find the side of the triangle

Connect OB and drop the perpendicular from point O to AB;

Because the center of the circumscribed circle lies at the intersection of the perpendiculars drawn from the midpoints of the sides, we get a right-angled triangle of the ВOD, ВOD is a right angle, the angle DВO = 30, since ВO – height, median and bisector;

The OD leg is equal to half of the hypotenuse:

ОВ = 1/2 * √3 / 6 = √3 / 12;

ВD ^ 2 + OD ^ 2 = OB ^ 2;

ВD ^ 2 + 3/144 = 3/36;

ВD ^ 2 = 1/12 – 1/48;

ВD ^ 2 = 4/48 – 1/48;

ВD ^ 2 = 3/48;

ВD ^ 2 = 1/16;

BD = 1/4;

AB = 2 * ВD;

AB = 1/4 * 2;

AB = 1/2;

Answer: the side of the triangle is 1/2 unit. rev.