The angle between the bisector and the height drawn from one vertex of an obtuse isosceles triangle is 48 degrees.

The angle between the bisector and the height drawn from one vertex of an obtuse isosceles triangle is 48 degrees. Determine the angles of the triangle

Let us designate the triangle given by the condition ABC, AB = BC. From the vertex of angle A we draw the bisector AM and the height AH on the continuation of the side BC beyond the vertex B.
By the condition of the problem, we know the angle HAM = 48 °.
In the right-angled triangle АНМ, we find the angle НМA:
∠ HMA = 90 ° – ∠ HAM = 90 ° – 48 ° = 42 °.
The variable x will denote the angle BAM, the angle B is equal to:
∠ B = 180 ° – (x + ∠ HMA) = 138 ° – x.
Angles A and C at the base of an isosceles triangle are 2x. Let’s make the equation:
2x + 2x + (138 ° – x) = 180 °
3x = 42 °
x = 14 °.
∠ A = ∠ C = 2 * 14 ° = 28 °.
∠ В = 138 ° – 14 ° = 124 °.
Answer: the angles of the triangle are 28 °, 28 °, 124 °.