In an isosceles triangle ABC with base AB, the angle C is 28. Find the angle between side AB

In an isosceles triangle ABC with base AB, the angle C is 28. Find the angle between side AB and the height AH of this triangle.

Since in an isosceles triangle ABC with a base AB ∠C = 28 °, then based on the fact that the sum of all the angles of the triangle is 180 ° and the angles at the base in an isosceles triangle are equal, ∠A = ∠ B = (180 ° – 28 °) / 2 = 76 °.

The height АН, lowered to the side ВС, leads to the formation of a right-angled triangle АНВ, in which ∠ В = 76 °, ∠Н = 90 ° is straight. Thus, the sought angle between the side AB and the height AH ∠BAN = 180 ° – 90 ° – 76 ° = 14 °.

Answer: ∠ВАН = 14 °.