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 an angle ABC = 177 degrees. Find the angle BOC.

Let’s immediately determine the magnitude of the angles in the ABC triangle:
since the sum of the angles of any triangle is 180, then we write the equality:
<BAC + <ABC + <BCA = 180 °,
<BAC + 177 ° + <BCA = 180 °, <BAC + <BCA = 180 ° – 177 ° = 3 °,
<BAC = <BCA = 3 ° / 2 = 1.5 °.
Next, consider <BOC: this angle is central for the BC arc, and <BAC rests on this arc, and by the inscribed angle theorem, the value of the inscribed angle is equal to half the central angle corresponding to the arc on which the inscribed angle rests. I.e:
<BOC = 2 * (<BAC) = 1.5 ° * 2 = 3 °.