In triangle ABC, angle C is 30 ° AD – bisector of angle A, angle B is 50 °, find the degree measure of angle B.

Let a triangle ABC be given, in which the degree measure of the angle ACB is 30 °, AD is the bisector of the angle BAC, the degree measure of the angle ADB is 50 °, then:

x is the degree measure of the CAD angle and the BAD angle, since ∠САD = ∠ВАD;

x + 30 ° is the degree measure of the external angle АDВ for the triangle АСD, since by the property of the external angles of the triangle ВАDВ = ∠САD + ∠АСD.

Knowing that ∠ADB = 50 °, we get the equation:

x + 30 ° = 50 °;

x = 20 ° – the degree measure of the CAD angle and the BAD angle;

2 ∙ 20 ° = 40 ° – the degree measure of the angle BAC;

180 ° – (40 ° + 50 °) = 90 ° is the degree measure of the ABC angle, since ∠ABS + ∠BAC + ∠ACB = 180 °.

Answer: The degree measure of the angle ABC is 90 °.