Find the cos of the angle of the triangle with vertices A (0; 6), B (1; 3), C (1; -1)
February 9, 2021 | education
| 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.
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