In an isosceles triangle ABC, the bisector of angle A intersects side BC at point K. Find the angles

In an isosceles triangle ABC, the bisector of angle A intersects side BC at point K. Find the angles of the triangle if AKB is 132 degrees.

The angles AKB and AKС are adjacent, the sum of which is 180, then the angle AKС = (180 – AKВ) = (180 – 132) = 48.
In an isosceles triangle, the angles at the base are equal, then the angle BAC = BCA.
Let the value of the angle BCA = 2 * X0, then the angle BAC = 2 * X0.
Since AK is the bisector of angle A, then the angle SAK = BAC / 2 = 2 * X / 2 = X0.
The sum of the inner angles of the AKС triangle is 1800, then:
(X + 2 * X + 48) = 180.
3 * X = 180 – 48 = 132.
X = 132/3 = 40.
Then the angle BAC = BCA = 2 * 44 = 88.
Angle ABC = (180 – 88 – 88) = 4.
Answer: The angles of the triangle are 88, 88, 4.