In an isosceles triangle, one of the corners is 140. Find the outer corner at the base of this triangle.

An isosceles triangle is a triangle in which the sides are equal.

Since in an isosceles triangle the angles at the base are equal, then:

∠А = ∠С.

Since the sum of all the angles of the triangle is 180º, then:

∠А = ∠С = (180º – ∠В) / 2;

∠А = ∠С = (180º – 140º) / 2 = 40º / 2 = 20º.

Since the sum of the outer and inner angles at one apex of the triangle is 180º, then:

φ = 180º – ∠А;

φ = 180º – 20º = 160º.

Answer: The outer angle at the base of the triangle is 160º.

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