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

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

Since the triangle ABC is isosceles, and the AC is its base, the angle BAC = BCA.

By condition, the sum of the angles (BAC + BCA) = 166.

Then the angle BAC = BCA = 166/2 = 83.

The sum of the inner angles of the triangle is 180, then the angle ABC = (180 – BAC – BCA) = (180 – 83 – 83) = 14.

The outer corners of a triangle are adjacent to the inner corners at the same vertex.

Then the ВAН angle = 180 – 83 = 97, the ВСD angle = 180 – 83 = 97, the СВE angle = 180 – 14 = 166.

Answer: The angles of the triangle ABC are 14, 83, 83, the outer angles are 97, 97, 166.