In triangle ABC, the outside angle at the apex A is 64º greater than the outside angle

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

Let x be the outer angle at the vertex B. Then the angle B = 180-x. According to the condition, the external angle at angle A is 64 degrees greater than the external angle at the vertex B and is equal to x + 64, then the angle A = 180- (x + 64) = 180-x-64 = 116-x. The sum of all the criminals of the triangle is 180 degrees, A + B + C = 180. Angle C = 80. We get the equation:
116’s + 180’s + 80 = 180;
116-2 * x = 0
x = 98-external angle at the apex of angle B, angle B = 180-x = 180-98 = 82.
Let’s check the external angle at the vertex A = x + 64 = 98 + 64 = 162, then the angle A = 180-162 = 18
18 + 82 + 80 = 180. Right.
Answer: angle B = 82