In a triangle ABC AB = AC, the outside angle at the base is 140 degrees. Find the angle AOC

In a triangle ABC AB = AC, the outside angle at the base is 140 degrees. Find the angle AOC formed at the intersection of the bisectors of angles C and A

The outer and inner corners of a triangle are adjacent corners. Find the inner corner C of a given isosceles triangle:
∠ C = 180 ° – 140 ° = 40 °.
The angles at the base are equal (an isosceles triangle).
Find the angle A at the vertex of this triangle:
∠ А = 180 ° – 2 * 40 ° = 100 °.
Consider the AOC triangle, in it:
∠ ОАС = 1/2 * ∠ А = 1/2 * 100 ° = 50 ° (AO is the bisector of angle A);
∠ OSA = 1/2 * ∠ C = 1/2 * 40 ° = 20 ° (CO is the bisector of angle C).
Two angles of the triangle are known, we find the third:
∠ AOC = 180 ° – (50 ° + 20 °) = 180 ° – 70 ° = 110 °.
Answer: The degree measure of the AOC angle is 110 °.