In a triangle ABC AC = BC. Angle C = 40 degrees. Find the outside corner at vertex B.

Since triangle ABC is isosceles, its angles at the base of AB are equal. Angle CAB = CBA = (180 – ACB) / 2 = (180 – 40) / 2 = 70.

Since the outer angle of the triangle is equal to the sum of the inner angles of the triangle that are not adjacent to it. Angle CBD = ACB + CAB = 40 + 70 = 110.

Answer: The outside angle CBD is 110.

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