Find the cosines of the angles of triangle ABC if A (1; 7), B (-2; 4), C (2: 0).

We are given a triangle ABC with the coordinates of points A (1; 7), B (-2; 4), C (2; 0). Let’s find the cosines of the angles of the triangle.

To do this, we apply the cosine theorem, but first of all we find the lengths of the sides of the triangle:

| AB | = √ ((- 2 – 1) ^ 2 + (4 – 7) ^ 2) = √18 = 3√2;

| BC | = √ ((2 – 1) ^ 2 + (7 – 0) ^ 2) = √50 = 5√2.

| AC | = √ ((- 2 – 2) ^ 2 + (4 – 0) ^ 2) = √32 = 4√2 /

By the cosine theorem, a ^ 2 = c ^ 2 + b ^ 2 – 2cb * cos α.

cos A = (AB ^ 2 + AC ^ 2 – CB ^ 2) / 2 * AB * AC = (18 + 50 – 32) / (2 * 5 * 3 * 2) = 36/50 = 3/5.

cos B = 0.

cos C = 4/5.