In an isosceles triangle ABC (AB- base), the angle ACB = 80 degrees. Find the outside angle at vertex B.

It is known that:

The ABC triangle is isosceles;
AB – base;
ACB angle = 80 °;
Find the outer angle at vertex B.

Decision.

Since the triangle is isosceles and AB is the base, then the sides of AC and BC are equal, that is, AC = BC. This means that their angles are also equal, that is, angle A = angle B.

In order to find the outer angle at the vertex B, you must first find the inner angle B of the triangle ABC.

Angle A = angle B = (180 ° – 80 °) / 2 = 100 ° / 2 = 50 °;

If the inside angle B = 50 °, then the outside angle at the apex B is 180 ° – 50 ° = 130 °.

Answer: 130 °.