The bisector AF and the height AH are drawn in an isosceles triangle ABC with base AC.

The bisector AF and the height AH are drawn in an isosceles triangle ABC with base AC. Find the angles of the triangle AHF if B = 112 degrees.

Consider the triangle ABН. Since AH is the height, then the angle AHB = 90. The angle ABH and ABC are adjacent angles, the sum of which is 180, then the angle ABH = 180 – 112 = 68. Then the angle BAН = 180 – 90 – 68 = 22.

Since the triangle ABC is isosceles, the angle BAC = BCA = (180 – ABC) / 2 = (180 – 112) / 2 = 68/2 = 34.

Since AF is the bisector of angle A, the angle BAF = BAC / 2 = 34/2 = 17.

Then the angle HAF = BAH + DFF = 22 + 17 = 39.

Then the angle AFB = 180 – 90 – 39 = 51.

Answer: The angles of triangle AHF are 39, 51, 90.