The height AH is drawn in an isosceles triangle ABC (AB = BC). Find the angle CAH if the angle B = 76 °.

Since, according to the condition, AB = BC, the triangle ABC is isosceles, and therefore the angles at the base of the AC are equal.

Angle BAC = BCA.

Since the sum of the inner angles of the triangle is 180, the sum of the angles BC and BCA is (180 – 76) = 104.

Then the angle BAC = BCA = 104/2 = 52.

Since AH is the height of the triangle, then the ACН triangle is rectangular, then the angle СAН = (180 – 90 – 52) = 38.

Answer: The СAН angle is 38.