The vertex of triangle ABC lies on a circle with Center O, angle A = 60, angle AOB: angle AOC = 3: 5.

The vertex of triangle ABC lies on a circle with Center O, angle A = 60, angle AOB: angle AOC = 3: 5. Find the unknown angles of the triangle.

Enter the proportionality factor:
∠ AOB = 3x;
∠ AOC = 5x.
In the isosceles triangle AOB, we find the angle at the base of the AOB:
∠ BAO = (180 ° – 3x) / 2;
Similarly, in the AOC triangle, we find the AO angle:
∠ CAO = (180 ° – 5x) / 2.
In the sum of these two angles give the angle A, which by condition is equal to 60 °.
(180 ° – 3x) / 2 + (180 ° – 5x) / 2 = 60 °
180 ° – 3x + 180 ° – 5x = 120
8x = 240
x = 30.
∠ AOB = 3x = 90 °;
∠ AOC = 5x = 150 °;
∠ BОС = 2 * ∠А = 120 °.
In an isosceles triangle BОС, the angles at the base are equal:
∠СВО = ∠ВСО = (180 ° – 120 °) / 2 = 30 °.
∠ В = ∠АВО + ∠СВО = 45 ° + 30 ° = 75 °
∠ С = ∠ВСО + ∠АСО = 30 ° + 15 ° = 45 °.
Answer: angle B = 75 °, angle C = 45 °.