Find the angle of an isosceles triangle opposite its base if its outer angle at the base is 130 degrees.

Let the isosceles triangle, given by the condition, be the triangle ABC, the angle BCD is the outer angle at the base of the AC.
Since the angle BCD is external, it is adjacent to the internal angle of the BCA. Adjacent angles add up to 180 degrees, because together they represent a flat angle (an angle that is 180 degrees).
Find the angle of the BCA:
BCA angle + BCD angle = 180 degrees;
BCA angle + 130 degrees = 180 degrees;
ICA angle = 180 degrees – 130 degrees;
ICA angle = 50 degrees.
Since the angle ABC is isosceles by the condition, and the AC is the base, the angles at its bases are equal. Then the angle BAC = angle BCA = 50 degrees.
The triangle sum theorem says that the sum of all the angles of any triangle is 180 degrees. It turns out:
angle BAC + angle ABC + angle BCA = 180 degrees.
Substitute the values ​​we know and find the degree measure of the angle ABC, which is opposite to the base of the AC:
50 degrees + angle ABC + 50 degrees = 180 degrees;
angle ABC = 180 degrees – 100 degrees;
angle ABC = 80 degrees.
Answer: angle ABC = 80 degrees.