The outer angle at the apex of an isosceles triangle is 4 times greater than the inner one.

The outer angle at the apex of an isosceles triangle is 4 times greater than the inner one. Find the outer corner at the base of the triangle.

Let an isosceles triangle △ ABC be given. The apex angle is B. And its outer angle is 4 * B. These angles add up to 180 degrees as defined by adjacent angles.

So B + 4 * B = 180 °; 5 * B = 180 °; B = 180 ° / 5 = 36 °.

Then the angles at the base of the triangle ABC, which are equal to each other, since the triangle is isosceles: <A = <C = (180 ° – 36 °) / 2 = 144 ° / 2 = 72 °.

Now define the adjacent angle at the base (180 – <A) = (180 ° – <C) = 180 ° – 72 ° = 108 °.