The ABC triangle is given by the coordinates of the vertices A (1; 3) B (2; 4) C (3; 3). Find the outer angle at vertex A.

For the solution, we use the formula for determining the length of a segment by the coordinates of points.

AB = √ (X2 – X1) ^ 2 + (Y2 – Y1) ^ 2 = √ (2 – 1) ^ 2 + (4 – 3) ^ 2 = √ (1 + 1) = √2 cm.

AC = √ (X2 – X1) ^ 2 + (Y2 – Y1) ^ 2 = √ (3 – 1) ^ 2 + (3 – 3) ^ 2 = √ (4 + 0) = 2 cm.

BC = √ (X2 – X1) ^ 2 + (Y2 – Y1) ^ 2 = √ (3 – 2) ^ 2 + (3 – 4) ^ 2 = √ (1 + 1) = √2 cm.

Since AB = BC, the triangle is isosceles.

The Pythagorean theorem holds in the triangle. AC ^ 2 = BC ^ 2 + AB ^ 2.

2 ^ 2 = (√2) ^ 2 + (√2) ^ 2.

4 = 4.

The triangle is rectangular, angle B = 90, then angle A = C = 45.

Then the outer angle at the vertex is A = 180 – 45 = 135.

Answer: The outside angle at vertex A is 135.