The vertices of triangle ABC are given. A (-6; -2) B (6; 7) C (4; -7) Find the inner corner A.

In order to find the angle A between the vector a, which is given by the points AB and the vector b, which is given by the points AC, it is necessary to use the formula for the angle between the vectors:

cos A = (a * b) / (| a | * | b |) or A = arccos ((a * b) / (| a | * | b |)).
Let’s find the coordinates of the vector a, given by points A and B:

AB = (6 – (-6); 7 – (-2)) = (12; 9).

Find the coordinates of the vector b given by points A and C:
AC = (4 – (-6); -7 – (-2)) = (10; -5).

Find the dot product a x b:

a x b = 12 * 10 + 5 * (-9) = 120 – 45 = 75.

Find the length of the vector a:

| a | = √ ((6 – (-6)) ^ 2 + (7 – (-2)) ^ 2)) = √ (12 ^ 2 + 9 ^ 2) = √225 = 15.

Let’s find the length of the vector b:

| b | = √ (((4 – (-6)) ^ 2 + (-7 – (-2)) ^ 2)) = √ (10 ^ 2 + 5 ^ 2) = √125 = 5√5.

Find the angle A between vectors a and b:
A = arcos ((75 / (15 * 5√5) = arcos (5 / 5√5) = arcos (1 / √5) ≈ arcos (0.45) ≈ 63 ° 26 ’.

Answer: angle А ≈ 63 ° 26 ‘.