In triangle ABC, the outside angle at vertex A is 64 degrees greater than the outside

In triangle ABC, the outside angle at vertex A is 64 degrees greater than the outside angle at vertex B. Find the angle B of triangle ABC if the angle C of triangle ABC is 80 degrees.

Let us take the value of the external angle at the vertex B as x degrees, then the value of the external angle at the vertex A is equal to (x + 64) degrees. The sum of the outer and inner angles at one vertex is 180 degrees. This means that the internal angle A is equal to (180 – x), and the internal angle at the vertex B is equal to (180 – (x + 64)) degrees. The sum of the angles of a triangle is (180 – x + 180 – (x + 64) + 80) degrees or 180 degrees. Let’s make an equation and solve it.

180 – x + 180 – (x + 64) + 80 = 180;

180 – x + 180 – x – 64 + 80 = 180;

-2x + 376 = 180;

-2x = 180 – 376;

-2x = – 196;

x = – 196: (- 2);

x = 98 – outer angle B;

180 – 98 = 82 – angle B of triangle ABC.

The answer is 82 degrees.