In a circle m center at point O, three radii OA, OB and OC of chords AB and BC

In a circle m center at point O, three radii OA, OB and OC of chords AB and BC are equal to the angle BOC 24 degrees, find the angles of the triangle AOB.

Since, by condition, the chord AB is equal to the chord BC, the arcs they contract are also equal, and therefore the central angles that rest on these arcs are also equal.

Angle AOB = BOC = 24.

The AOB triangle is isosceles, since AO = BO = R, then in the AOB triangle the angle OAB = OBA.

Angle ОАВ = ОВА = (180 – AOB) / 2 = (180 – 24) / 2 = 156/2 = 78.

Answer: The angles of the AOB triangle are 24, 78, 78.