Angles B and C of triangle ABC are equal to 10º and 100º respectively.

Angles B and C of triangle ABC are equal to 10º and 100º respectively. Find the angles BOC, COA, where O is the center of the circumscribed circle.

Find the third corner of the triangle, angle A.
∠A = 180 ° – (∠B + ∠C) = 180 ° – 110 ° = 70 °.
Angle A is inscribed in a given circle according to the condition and rests on the arc BC, the central angle of the VOC also rests on the same arc, therefore:
∠ BOC = 2 * ∠A = 140 °.
We argue in a similar way about the COA angle. It is central, rests on the same arc AC, on which the inscribed angle B rests.
∠ СОА = 2 * ∠В = 20 °.
Answer: BOS angle 140 °, SOA angle 20 °.