In an isosceles triangle ABC with a base AC, the lateral side and the angle at the base AB = 9, C = 15 degrees are known.

In an isosceles triangle ABC with a base AC, the lateral side and the angle at the base AB = 9, C = 15 degrees are known. Find the length of the CH elevation drawn to the AB side.

Either you misread the assignment, or you misunderstood something. Angle C cannot have a degree measure of 15, because the triangle is isosceles, the angle A will also remain with the degree measure 15. 15 + 15 = 30 degrees. Accordingly, the angle above the base = 150 degrees. And this cannot be, because the height of СН, drawn to AB, forms a right angle in the СНВ triangle, which contradicts the laws of mathematics. The sum of the angles of a triangle is always 180 degrees, but here only two angles gave 240 degrees.