In an isosceles triangle ABC, the angle at the vertex B is 120 degrees, AC = 2 roots of 21. Find the length of the median AM

Given: ΔABC – isosceles, AM – median.
∠B = 120 °, AC = 2√21.
Find: AM.
Decision:
ΔABC – isosceles => ∠A = ∠С = (180 ° – 120 °) ÷ 2 = 30 °.
By the sine theorem:
AC: sin (∠B) = BC: sin (∠A);
2√21: sin (120 °) = BC: sin (30 °);
BC = (2√21 × 0.5): (√3) / 2;
BC = 2√7.
AM – median => BM = MC = 0.5 × BC = √7.
By the cosine theorem:
AM ^ 2 = MC ^ 2 + AC ^ 2 – 2 × cos (∠C) × MC × AC;
AM ^ 2 = 7 + 84 – 2 × (√3) / 2 × √7 × 2√21;
AM ^ 2 = 91 – 42;
AM ^ 2 = 49;
AM = 7.
Answer: AM = 7.