Find the side and area of an isosceles triangle if the base is 18 cm and the angle opposite to the base is 120.

1. Apply the cosine theorem to an isosceles triangle ABC with base AC:

AC = 18;
AB = BC = x;
∠B = 120 °.
AC ^ 2 = AB ^ 2 + BC ^ 2 – 2 * AB * BC * cos∠B;
18 ^ 2 = x ^ 2 + x ^ 2 – 2x ^ 2 * cos120 °;
324 = 2x ^ 2 – 2x ^ 2 * cos (180 ° – 60 °);
324 = 2x ^ 2 + 2x ^ 2 * cos60 °;
324 = 2x ^ 2 + 2x ^ 2 * 1/2;
324 = 3x ^ 2;
x ^ 2 = 324/3;
x = 18 / √3 = 6√3.
2. We find the height h by the Pythagorean theorem:

h ^ 2 + 9 ^ 2 = x ^ 2;
h ^ 2 = x ^ 2 – 81 = (6√3) ^ 2 – 81 = 108 – 81 = 27;
h = √27 = 3√3.
3. Area of a triangle:

S = 1/2 * h * AC;
S = 1/2 * 3√3 * 18 = 27√3.
Answer: 6√3 cm; 27√3 cm ^ 2.