The angle between the height and the side of an isosceles triangle is 15 degrees

The angle between the height and the side of an isosceles triangle is 15 degrees less than the angle at its base, find the angles of the triangle.

Given △ ABC: AB = BC, ∠A = ∠C, BH – both the height, and the median, and the bisector drawn from the top of an isosceles triangle to its base.
1. Consider △ AHB: ∠AHB = 90 ° (since BH is height).
Let us denote ∠BAH (aka ∠A) as x, then ∠ABH (the angle between the height BH and side AB) is x – 15 °.
By the theorem on the sum of the angles of a triangle:
∠AHB + ∠BAH + ∠ABH = 180 °;
90 ° + x + x – 15 ° = 180 °;
2 * x = 180 ° – 90 ° + 15 °;
2 * x = 105 °;
x = 105 ° / 2;
x = 52.5 °.
Thus, ∠A = ∠C = 52.5 °.
2. Since BH is a bisector, then ∠ABH = ∠CBH = ∠B / 2.
In this way:
x – 15 ° = ∠B / 2;
∠B = 2 * (x – 15 °) = 2 * (52.5 ° – 15 °) = 2 * 37.5 ° = 75 °.
Answer: ∠A = ∠C = 52.5 °, ∠B = 75 °.