The angle between the height and the bisector drawn from one vertex of an obtuse isosceles

The angle between the height and the bisector drawn from one vertex of an obtuse isosceles triangle is 36 degrees. Determine the angles of the triangle.

We denote the triangle given by the condition ABC, angle B is the top, angles A and C are angles at the base.
Draw the AM bisector to the BC side.
The triangle is obtuse, the height AH from the vertex A will intersect with the side BC on its extension beyond the vertex B.
Angle MAН = 36 ° (as required).
In the right-angled triangle НAM we find the second acute angle НMA:
∠ HMA = 90 ° – ∠ MAН = 90 ° – 36 ° = 54 °.
In the YAM triangle, we denote the BAM angle by x, then:
∠ B = 180 ° – (x + ∠ HMA) = 126 ° – x.
∠ A = ∠ C = 2 * ∠ BAM = 2x.
Let’s make the equation:
(126 ° – x) + 2x + 2x = 180 °
3x = 54 °
x = 18 °.
∠ A = ∠ C = 36 °.
∠ B = 108 °.
Answer: the angles of the triangle are 36 °, 36 °, 108 °.