In triangle ABC, the height dropped from the vertex B intersects the AC side at point H, and the bisector of angle B

In triangle ABC, the height dropped from the vertex B intersects the AC side at point H, and the bisector of angle B intersects AC at point M, angle ABH = 23 degrees, angle BMA = 64 degrees. Find the angles of the triangle ABC.

Since BH is the height of a triangle, triangles ABH and BHM are rectangular.

Then the angle НВМ = (180 – ВНМ – ВМН) = (180 – 90 – 64) = 26.

Angle BAN = (180 – AHB – ABH) = (180 – 90 – 23) = 67.

Angle ABM = (ABH + HBM) = (23 + 26) = 49.

Since BM, by condition, is the bisector of the angle ABC, then the angle ABC = ABM * 2 = 49 * 2 = 98.

The sum of the inner angles of the triangle is 180, then the angle ACB = (180 – BAC – ABC) = (180 – 67 – 98) = 15.

Answer: The angles of the triangle ABC are equal to 67, 98, 15.