Find the angle ABC of a triangle with vertices A (30; 43), B (26; 45), C (25; 48).

We need to find the value of the angle ABC of the triangle with the vertices A (30; 43), B (26; 45), C (25; 48).

Let’s start the solution by finding the coordinates of the vector BA (4, -2) and BC (-1, 3) (we subtracted the coordinates of the start from the coordinates of the end of the segment).

The next step is to find the length of these vectors:

| BA | = √ (4 ^ 2 + (-2) ^ 2) = √ (16 + 4) = √20 = 2√5.

| BC | = √ ((- 1) ^ 2 + 3 ^ 2) = √ (1 + 9) = √10.

Let’s apply the formula to calculate the cosine of the angle ABC:

cos ABC = (BA * BC) / | BA | * | BC | = (4 * (-1) + (-2) * 3) / 2√5 * √10 = -10 / 10√2 = -1 / √2.

angle ABC = 3p / 4 = 135 °.