Triangle ABC is given by the coordinates of its vertices: A (-1; 1), B (0; 2), C (1; 1). Find the outside angle at vertex A.

Let’s find the lengths of the sides AB and BC of a triangle using the formula to find the length of a vector.

So, the lengths of the sides are equal:

AB = √ ((- 1 – 0) ^ 2 + (1 – 2) ^ 2) = √2;

BC = √ ((0 – 1) ^ 2 + (2 – 1) ^ 2) = √2.

We conclude that the triangle is isosceles, since AB = BC.

If you draw a triangle on the coordinate plane, then you can see that point B lies on the OY axis, which means that triangle ABO is right-angled.

Angle ABO = 45 ° and angle A = angle C.

It remains to find the outer angle at the vertex A:

180 ° – 45 ° = 135 °.