In triangle ABC, angle B is 45 ‘, and angle C is 15’ less than angle B. Find the outer angle at the vertex A.

Let us determine the value of the angle ACB, which, by condition, is 15 less than the angle ABC.

Angle ACB = 45 – 15 = 30.

Then the angle BAC = (180 – 45 – 30) = 105.

The sought angle CAD is adjacent to the angles of BAC, then the angle of CAD = (180 – BAC) = (180 – 105) = 75.

Second way.

The angle BAC, AO condition, is equal to: BAC = (45 – 15) = 30.

The external angle CAD is equal to the sum of the two internal angles of the triangle that are not adjacent to it.

SAD angle = (ABC + BAC) = 45 + 30 = 75.

Answer: The outside angle at vertex A is 75.