In triangle ABC, AD is a bisector, CAD angle is 26 degrees, ADB angle is 110 degrees.

In triangle ABC, AD is a bisector, CAD angle is 26 degrees, ADB angle is 110 degrees. Find the outside corner at vertex B.

Consider a triangle ABD in which the angles are known:
∠ ADB = 110 ° (according to the problem statement);
∠ BAD = ∠ CAD = 26 ° (CD – bisector by condition).
Find the angle ABD:
∠ ABD = 180 ° – (∠ ADB + ∠ BAD) = 180 ° – 136 ° = 44 °.
The outer and inner corners of the triangle are adjacent, we find the outer corner B.
∠ In ext. = 180 ° – ∠ ABD = 180 ° – 44 ° = 136 °.
It was possible to use the property of the outer corner of the triangle:
∠ In ext. = ∠ ADB + ∠ BAD = 110 ° + 26 ° = 136 °.
Answer: the degree measure of the external angle at the vertex B is 136 °.