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.

The sum of all the angles of a triangle, regardless of its type, is 180º. Since in an isosceles triangle the angles at the base are equal, then an obtuse angle in it can only be the angle opposite to the base, that is, the angle ∠B.

Thus:

∠А = ∠С = (180º – ∠В) / 2;

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

The outside angle at the apex of a triangle is the angle adjacent to the apex of the triangle. Together they make up an unfolded angle, the degree measure of which is 180º.

Based on this:

φ = 180º – ∠А

φ = 180º – 20º = 160º.

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




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