Find the angles of an isosceles triangle if the angle at the base is 3 times less than the outer corner adjacent to it.

An isosceles triangle is a triangle in which opposite sides are equal and the angles at the base are equal:

AB = BC;

∠А = ∠С.

Since the sum of the outer and inner angles of the triangle at one vertex is equal to 180º, and the value of the inner angle is three times less than the value of the outer one, we express:

x is the degree measure of the internal angle at the vertices ∠А and ∠С;

3x – the degree measure of the external angle at the vertices ∠А and ∠С;

x + 3x = 180;

4x = 180;

x = 180/4 = 45;

∠А = ∠С = 45º.

Since the sum of all the angles of the triangle is 180º, and the angles ∠А and ∠С at the base are equal to 45º, then:

∠В = 180º – ∠А – ∠С;

∠В = 180º – 45º – 45º = 90º.

Answer: ∠А = ∠С = 45º; ∠В = 90º.