In an isosceles triangle ABC with base AC, the angle A is 66 °. Find the outside angle at vertex B.

First way.

In an isosceles triangle, the angles at its base are equal, then the angle ACB = BAC = 66. Since the sum of the inner angles of the triangle is 180, then the angle ABC = 180 – BAC – BCA = 180 – 66 – 66 = 48. Then the external angle AВD = 180 – 48 = 132.

Second way.

The outer corner of a triangle is equal to the sum of two inner angles not adjacent to it

AВD angle = BAC + BCA = 66 + 66 = 132.

Answer: The outside angle at vertex B is 132.

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