The lateral side of an isosceles triangle is 4. The angle at the apex opposite to the base

The lateral side of an isosceles triangle is 4. The angle at the apex opposite to the base is 120 degrees. Find the diameter of the circle around this triangle.

ABC – isosceles triangle, AB = BC = 4, ∠ABС = 120 °.
1. By the theorem on the sum of the angles of a triangle:
∠А + ∠В + ∠С = 180 °.
∠A = ∠C = x – since these are the angles at the base of an isosceles triangle, ∠B = ∠ABС = 120 °. Then:
x + 120 ° + x = 180 °;
2x = 180 ° – 120 °;
2x = 60 °;
x = 60 ° / 2;
x = 30 °.
∠А = ∠С = x = 30 °.
2. From the top B to the base of the AC, draw the height BH. Since △ ABC is isosceles, the BH is the height, the median, and the bisector.
Consider △ AНВ: ∠AНВ = 90 °, ∠HAB = 30 °, AB = 4 – hypotenuse.
In a right-angled triangle, opposite an angle of 30 °, lies a leg, which is 2 times less than the hypotenuse. Then:
BH = AB / 2;
BH = 4/2;
BH = 2.
3. The radius of a circle circumscribed about an isosceles triangle is found by the formula:
R = b² / 2h,
where b is the length of the side, h is the height drawn to the base.
Find the radius:
R = 4² / 2 * 2 = 16/4 = 4.
4. The diameter of the circle is:
D = 2R;
D = 2 * 4 = 8.
Answer: D = 8.