Find the cos of the angle of the triangle with vertices A (0; 6), B (1; 3), C (1; -1)

We are given a triangle with its vertices A (0; 6), B (1; 3) and C (1; -1). And we need to find the cosines of the angles of the triangle.

And we start by calculating the lengths of the sides of the triangle.

AB = √ (9 + 9) = 3√2;

BC = √ (16 +16) = 4√2;

AC = √ (1 + 49) = 5√2.

Next, we will apply the Cosine Theorem and calculate the cosine of each corner of the triangle.

Find the cosine of the angle B:

50 = 18 + 32 – 12 * 4 * cos B;

cos B = 0, so

angle B = 90 °.

We calculate the cosine of the angle C:

18 = 50 + 32 – 80 cos C;

cos C = (82 – 18) / 80 = 64/80 = 0.8;

Cosine A:

32 = 18 + 50 – 60 * cos A;

cos A = 18 * 2/60 = 0.6.