The angle at the base of an isosceles triangle ABC is 32 degrees, AB is its lateral side

The angle at the base of an isosceles triangle ABC is 32 degrees, AB is its lateral side, AM is the bisector of the triangle. Find the angles of triangle ABM.

In the first case AB = AC, then the bisector AM will also be the height of the triangle.

Then the triangle AFM is rectangular, angle AMC = 900, angle AFM = 32, angle CAM = 180 – 90 – 32 = 58.

Answer: The angles of the triangle ABM are 90, 32, 58.

In the second case, BA = BC. Angle BAM = CAM = BAC / 2 = 32/2 = 160.

Angle AMC = 180 – CAM – BCM = 180 – 16 – 32 = 132.

The angles ВМА and СМА are adjacent, then the angle АМВ = 180 – АМС = 180 – 132 = 48.

Then the angle ABM = 180 – BAM – AMB = 180 – 16 – 48 = 116.

Answer: The angles of the triangle ABM are 16, 48, 116.