In an isosceles triangle ABC with base AC, angle A + angle C = 166 degrees.

In an isosceles triangle ABC with base AC, angle A + angle C = 166 degrees. Find all the inside and outside corners of a triangle ABC.

Angle A and angle C are angles at the base, respectively, they are equal.
Angle A = Angle B = 166 ° / 2 = 83 °
Find the angle at the top, angle B:
Angle B = 180 ° – 166 ° = 14 °.
Find the outer angle at the vertex B:
180 ° – 14 ° = 166 °.
The outer angles at the vertices of the triangle at the base are equal:
180 ° – 83 ° = 97 °.
Answer: The inner angles of the triangle are 83 °, 83 °, 14 °; outer angles of the triangle: 97 °, 97 °, 166 °.