In an isosceles triangle ABC, the angle at the base of the BAC is 50 °.

In an isosceles triangle ABC, the angle at the base of the BAC is 50 °. Find the angle between the height AH of the triangle and its lateral side AB.

Given:

ABC – isosceles triangle;

BAC = 50 °;

AH – height;

Find: the angle between AH and the side AB (BAH).

Consider an isosceles triangle ABC.

According to one of the properties of an isosceles triangle, the angles at its base AC are equal, i.e. BAC = BCA = 50 °.

Hence, according to the property of the sum of the angles of a triangle, the angle ABC = 180 ° – BAC – BCA = 180 ° – 2 * 50 ° = 180 ° – 100 ° = 80 °.

Next, consider the triangle ABH, it is rectangular, since AH is the height. Those. angle BHA = 90 °. Hence, according to the same property of the sum of the angles of a triangle, the angle BAH = 180 ° – BHA – ABH = 180 ° – 90 ° – 80 ° = 10 °.

Answer: The angle between height AH and side AB is 10 °.