One of the outer corners of a triangle is twice the size of the other outer corner of that triangle.

One of the outer corners of a triangle is twice the size of the other outer corner of that triangle. Find the smaller of them if the inner corner of the triangle, not adjacent to the indicated outer corners, is 60 degrees.

Let a triangle ABC be given: angle C = 60 degrees, external angle at apex A – angle КAС, external angle at vertex B – СВН.
1. The angles КAС and CAB (angle A) are adjacent. The sum of adjacent angles is 180 degrees. Let’s designate the angle KAС as x, then:
CAS angle + CAB angle = 180 degrees;
x + angle CAB = 180;
angle CAB = 180 – x.
Angle A = angle CAB = 180 – x.
2. Angles СВН and ABC (angle B) are adjacent. The sum of adjacent angles is 180 degrees. Let’s designate the angle of the СВН as y, then:
СВН angle + ABC angle = 180 degrees;
y + angle ABC = 180;
angle ABC = 180 – y.
Angle B = angle ABC = 180 – y.
3. It is known from the condition that one of the outer corners of a triangle is twice as large as the other outer corner of this triangle. Let the external angle at the vertex B be 2 times greater than the external angle at the vertex A. Then:
СВН angle = 2 * KAС angle;
y = 2x.
4. By the theorem on the sum of the angles of a triangle:
angle A + angle B + angle C = 180 degrees;
180 – x + 180 – y + 60 = 180.
Together, we substitute 2x for the values ​​of y:
180 – x + 180 – 2x + 60 = 180;
-3x = -240;
x = 240/3;
x = 80.
CAS angle = x = 80 degrees.
СВН angle = y = 2x = 2 * 80 = 160 (degrees).
Answer: КАС angle = 80 degrees.