# The angle at the apex of an isosceles triangle opposite to the base is 120 °. At what angle

**The angle at the apex of an isosceles triangle opposite to the base is 120 °. At what angle is each side of the triangle visible from the center of the circumscribed circle around it?**

Since triangle ABC is isosceles, its angle BAC = ABC = (180 – ASB) / 2 = (180 – 120) / 2 = 60/2 = 30.

The inscribed angle ABC rests on the arc AC, then the degree measure of the arc AC = 2 * 30 = 60.

The AC side from the top O of the circle is visible at the central angle AOC, the degree measure of which is equal to the arc AC. Angle AOC = 60.

The inscribed angle BAC rests on the arc BC, then the degree measure of the arc BC = 2 * 30 = 60.

The BC side from the top O of the circle is visible at the central angle BOC, the degree measure of which is equal to the BC arc. BOC angle = 60.

The inscribed angle ACB is 120 and rests on a large arc AB, the degree measure of which is 120 * 2 = 240. Then the arc ACB, on which the side AB of the triangle rests, is 360 – 240 = 120.

The central angle AOB, at which side AB is seen, is equal to the arc ACB. Angle AOB = 120.

Answer: The sides of the triangle are visible from the center of the circle at angles 30, 30, 120.