A circle centered at point O is described around an isosceles triangle ABC, in which AB = BC

A circle centered at point O is described around an isosceles triangle ABC, in which AB = BC and angle ABC = 25 degrees. Find the BOC angle.

The triangle ABC, AO condition is isosceles, then the angle BAC = BCA = (180 – ABC) / 2 = (180 – 25) / 2 = 155/2 = 77.5.

The inscribed angle BAC rests on the BC arc, then the degree measure of the BC arc is equal to the degree measure of the BAC angle. Arc BC = 77.5.

The central angle BOC is also based on the arc BC, then the angle BOC is equal to two degree measures of the arc BC.

Angle BОС = 2 * ВС = 2 * 77.5 = 155

Answer: The BOC angle is 155.