In triangle ABC, angle B = 60 °. Outside angle at apex A = 120 °, CH- bisector to side AB.

In triangle ABC, angle B = 60 °. Outside angle at apex A = 120 °, CH- bisector to side AB. Find angle A, side AH, if segment AB = 18 cm.

The outer corner CAD and the inner corner BAC are adjacent angles, the sum of which is 180, then the angle BAC = (180 – 120) = 60.

In triangle ABC, two angles are equal to 60, then the third angle ACB = (180 – 60 – 60) = 60.

Then the triangle ABC is isosceles, and therefore, the bisector CH is also the median of the triangle, then AH = AB / 2 = 9 cm.

Answer: The angle BAC is 60, the segment AH is 9 cm.