Find the cosine of the angle A of triangle ABC if A (3; 9), B (0; 6), C (4; 2)

First of all, let’s calculate the lengths of the sides of this triangle ABC. For this purpose, we will use the formula for calculating the distance between two points. We have: AB = √ (0 – 3) ² + (6 – 9) ²) = √ (9 + 9) = 3√ (2); AC = √ (4 – 3) ² + (2 – 9) ²) = √ (1 + 49) = 5√ (2) and AC = √ (4 – 0) ² + (2 – 6) ²) = √ (16 + 16) = 4√ (2).
Now we will use the following formula of the cosine theorem BC² = AB² + AC² – 2 * AB * AC * cos∠A, which we rewrite as cos∠A = (AB² + AC² – BC²) / (2 * AB * AC). We have cos∠A = ((3√ (2)) ² + (5√ (2)) ² – (4√ (2)) ²) / (2 * 3√ (2) * 5√ (2)) = (18 + 50 – 32) / (2 * 15 * 2) = 36/60 = 0.6.
Answer: 0.6.