The center of a circle circumscribed about a triangle belongs to one of the sides of the triangle, one of the angles

The center of a circle circumscribed about a triangle belongs to one of the sides of the triangle, one of the angles of which is 34 degrees. Find the rest of the corners of the triangle.

The center of the circumscribed circle O lies on the side AB of the triangle ABC. Let’s connect the points O and C with a segment. It is equal to the segments AO and OB, since AO, OB and OС are the radii of the circle. Then the triangles AOC and BOC are isosceles, which means that ∠ОВС = ∠ОСВ and ∠ОАС = ∠OСA.

Since the sum of the angles of the triangle is 180 °, we have:

∠ОВС + ∠ОАС + ∠АСВ = 180 °, while ∠АСВ = ∠ОСА + ∠ОСВ = ∠ОАС + ∠ОВС.

Consequently:

2 ⋅ (∠ОВС + ∠ОАС) = 180 °.

Hence ∠АСВ = ∠ОВС + ∠ОАС = 90 °.

Then the angle 34 ° can be either OBC or OAC. In any case, the remaining angle is:

90 ° – 34 ° = 56 °.

Answer: the remaining angles are 56 ° and 90 °.