Find the angle A in a triangle with vertices A (1.2√3), B (-1.0), C (1.0).
| April 4, 2021 | education
By the coordinates of the vertices of the triangle ABC, we determine the lengths of its sides.
AB = √ (X1 – X2) ^ 2 + (Y1- Y2) ^ 2 = √ (-1 – 1) ^ 2 + (0 – 2 * √3) ^ 2 = √ (4 + 12) = √16 = 4 cm.
AC = √ (X1 – X2) ^ 2 + (Y1- Y2) ^ 2 = √ (1 – 1) ^ 2 + (0 – 2 * √3) ^ 2 = √12 = 2 * √3 cm
BC = √ (X1 – X2) ^ 2 + (Y1- Y2) ^ 2 = √ (1 – (-1)) ^ 2 + (0 – 0) ^ 2 = √4 = 2 cm.
By the cosine theorem, we determine the value of the angle A.
Side BC lies opposite corner A, then:
BC ^ 2 = AB ^ 2 + AC ^ 2 – 2 * AB * AC * CosA.
4 = 16 + 12 – 2 * 4 * 2 * √3 * CosA.
16 * √3 * CosA = 24.
CosA = 24/16 * √3 = 24 * √3 / 48 = √3 / 2,
Angle A = arcos (√3 / 2) = 300.
Answer: Angle A is 300.
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