In an isosceles triangle ABC, the angle at its base AC is equal to a, the segment BK is the height of triangle ABC.

In an isosceles triangle ABC, the angle at its base AC is equal to a, the segment BK is the height of triangle ABC. Find the degree measures of the angles of the triangle BKC.

Since triangle ABC is isosceles, its height BK is also its median and bisector of angle ABC, then the angle СBK = ABK = (α / 2) ^ 0.

The BCK triangle is rectangular, since the BC is height, then the angle BCK = (180 – BKC- СBK) = (180 – 90 – α / 2) = (90 – α / 2) ^ 0.

Answer: The angles of the triangle BSC are 90, (α / 2) ^ 0, (90 – α / 2) ^ 0