Find the angles of a triangle ABC if A (-6; -7), B (6; 7), C (2; -1)

We have the coordinates of the vertices:

A (-6; -7), B (6; 7), C (2; -1).

Let’s find the lengths of the sides of the triangle:

| AB | = ((6 + 6) ^ 2 + (7 + 7) ^ 2) ^ (1/2) = (36 + 49) ^ (1/2) = 85 ^ (1/2);

| AC | = ((2 + 6) ^ 2 + (-1 + 7) ^ 2) ^ (1/2) = (64 + 36) ^ (1/2) = 10;

| BC | = ((2 – 6) ^ 2 + (-1 – 7) ^ 2) ^ (1/2) = (16 + 64) ^ (1/2) = 78 ^ (1/2);

AB ^ 2 = AC ^ 2 + BC ^ 2 – 2 * | AC | * | BC | * cos C;

85 = 100 + 78 – 2 * 10 * 78 ^ (1/2) * cos C;

85 = 178 – 20 * 8.83 * cos C;

93 = 176.6 * cos C;

cos C = 0.53;

C = 58 °;

AC ^ 2 = AB ^ 2 + BC ^ 2 – 2 * | AB | * | BC | * cos B;

100 = 85 + 78 – 2 * 81.42 * cos B;

162.84 * cos B = 63;

cos B = 0.387;

B = 67 °;

BC ^ 2 = AB ^ 2 + AC ^ 2 – 2 * | AB | * | AC | * cos C;

78 = 100 + 85 – 2 * 10 * 85 ^ (1/2) * cos C;

184.4 * cos C = 107;

cos C = 0.58;

C = 55 °.