Question: ABC is an equilateral triangle with side a. The triangle ABC is circumscribed about a circle with center O. If the length of the side of the
equilateral triangle is a, then what is the radius of the circle?

Solution:

Because ABC is an equilateral triangle, so

angle B = angle C = angle A = 60^{o}

Because AD perpendicular to BC at point D, so triangle BAD is a right triangle.

sin B = AD/AB, so

AD = AB sin B = a sin60^{o} = (square root of 3) a/2

Because BE perpendicular to AC at point E, and ABC is an equilateral triangle, so

BE cuts angle B in half, so angle OBD = 30^{o}

Because OD is a radius and angle OBD = 30^{o}, so

OB = 2r (In a right triangle, the leg opposite 30^{o} is one-half the hypotenuse.)

BE = BO + OE = 2r + r = 3r

Because in an equilateral triangle, the three altitude are equal, so