In triangle ABC, angle B = 120, AB = 7cm, AC = 13cm. Find the perimeter and area of triangle ABC.

By the cosine theorem, we calculate the length of the BC side.

AC ^ 2 = AB ^ 2 + BC ^ 2 – 2 * AB * BC * Cos120.

169 = 49 + BC ^ 2 – 2 * 7 * BC * (-1/2).

BC ^ 2 + 7 * BC – 120 = 0.

Let’s solve the quadratic equation.

D = b ^ 2 – 4 * a * c = 72 – 4 * 1 * (-120) = 49 + 480 = 529.

BC1 = (-7 – √529) / 2 * 1 = (-7 – 23) / 2 = -30 / 2 = -15. (Doesn’t fit because <0).

BC2 = (-7 + √529) / 2 * 1 = (-7 + 23) / 2 = 16/2 = 8.

BC = 8 cm.

Determine the perimeter of the triangle. Ravs = 7 + 13 + 8 = 28 cm.

Let’s calculate the area of the triangle.

Savs = AB * BC * Sin120 = 7 * 8 * √3 / 2 = 28 * √3 cm2.

Answer: The perimeter of the triangle is 28 cm, the area is 28 * √3 cm2.