In an isosceles triangle ABC (AB = BC), the bisector of angle A intersects the side BC

In an isosceles triangle ABC (AB = BC), the bisector of angle A intersects the side BC at point K. Find the angles of triangle ABC if AKB = 132.

The AKC angle is adjacent to the AKB angle, the sum of which is 180.

Then the angle AKC = (180 – AKВ) = (180 – 132) = 48.

Let the angle ACB = X0.

Since the triangle ABC is isosceles, the angle CAB = ACB = X0.

AK = bisector of the angle BAC, then the angle СAK = BAC / 2 = X / 2.

The sum of the interior angles of a triangle is 180.

Then in the triangle ACK, 180 = 48 + X + X / 2.

3 * X / 2 = 132.

X = 132 * 2/3 = 88.

Then the angle ABC = 180 – 88 – 88 = 4.

Answer: The angles of the triangle ABC are equal to 40, 88, 88.