In triangle ABC AB = 23m, BC = 7m, AC = 24m. Find the angle A, the area of the triangle.

By Heron’s theorem, we determine the area of the triangle ABC.

The semi-perimeter of the triangle is: p = (AB + BC + AC) / 2 = (23 + 7 + 24) / 2 = 27 cm.

Then Sav = √27 * (27 – 23) * (27 – 24) * (27 – 7) = √6480 = 36 * √5 cm2.

The value of the angle BAC is determined by the cosine theorem for a triangle.

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

49 = 529 + 576 – 2 * 23 * 24 * CosA.

1104 * CosA = 1056.

CosA = 1056/1104 = 22/23.

Angle BAC = arcos (22/23).

Answer: The area of the triangle is 36 * √5 cm2, the angle A is equal to arcos (22/23).