You are given the vertices of the triangle A (-6; 3; 7) B (-4; 3 ;; 5) C (-1; 8; 7). find the angle BAC of this triangle.

Find the lengths of the sides of the triangle ABC:

| AB | = ((-4 + 6) ^ 2 + (3 – 3) ^ 2 + (5 – 7) ^ 2) ^ (1/2) = 8 ^ (1/2);

| AC | = ((-1 + 6) ^ 2 + (8 – 3) ^ 2 + (7 – 7) ^ 2) ^ (1/2) = (25 + 25) ^ (1/2) = 50 ^ (1 / 2);

| BC | = ((-1 + 4) ^ 2 + (8 – 3) ^ 2 + (7 – 5) ^ 2) ^ (1/2) = (9 + 25 + 4) ^ (1/2) = 38 ^ (1/2);

The value of the angle A is found by the cosine theorem:

BC ^ 2 = AC ^ 2 + AB ^ 2 – 2 * AC * AB * cos A;

38 = 50 + 8 – 2 * (50 * 8) ^ (1/2) * cos A;

-20 = -2 * 20 * cos A;

-20 = -40 * cos A;

cos A = 1/2;

A = 60 °.