In an isosceles triangle ABC, the ABC angle is 50 degrees. what are the angles BAC and ACB equal to?

Let’s take advantage of the fact that the sum of the angles of any triangle is 180 °.

According to the condition of the problem, the angle ABC of the isosceles triangle ABC is 50 °.

Consider 2 cases:

1) the angle ABC is the angle at the base of the given triangle ABC.

Then the second angle at the base of this triangle is also 50 °.

Let this be the BAC angle.

Since the sum of the angles of the triangle is 180 °, the angle ACB is 180 – 50 – 50 = 180 – 100 = 80 °.

Thus, ∠BAC = 50 °, ∠ACB = 80 °.

2) the angle ABC is the angle opposite to the base of the given triangle ABC.

Then ∠BAC = ∠ACB, since these angles lie at the base of an isosceles triangle.

Denoting the value of the angle ∠BAC through x, we can compose the following equation:

x + x + 50 = 180.

We solve this equation:

2x + 50 = 180;

2x = 180 – 50;

2x = 130;

x = 130/2;

x = 65.

Therefore: ∠ВАС = ∠АСВ = 65 °.