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.
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