In an isosceles triangle ABC with base AC, the bisector BN of the outer angle at the vertex B is drawn. Determine the angles CBN if the angle A = 71 degrees
The external angle of a triangle is equal to the sum of the two internal angles of the triangle that are not adjacent to it.
Angle CBD = (BAC + BCA).
Since the triangle ABC is isosceles, the angle BCA = BAC = 71, then the angle CBD = 71 + 71 = 142.
ВН is the bisector of the ВСD angle, then the angle DВН = СBН = CBD / 2 = 142/2 = 71.
Answer: The value of the СBН angle is 71
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