One of the base angles of an isosceles triangle is 40 degrees less than the outer base angle.

One of the base angles of an isosceles triangle is 40 degrees less than the outer base angle. Find the outer corner.

Let’s denote the vertices of the triangle as a, b, and c. And the angles at the base will be <a = <c, and the angle <b is the angle at the apex. The outer angle is with any of the angles at the base, <a or <c in the sum of 180 degrees, and the angle itself is 40 degrees less. Let’s write it like this:

<c + (<c + 40 °) = 180 °; 2 * <c = 180 ° – 40 ° = 140 °.

Then <c = 140 ° / 2 = 70 °.

Let’s define an external angle with an angle <c.

The outer desired angle is (40 ° + <c) = 70 ° + 40 ° = 110 °.