Bisector BD is drawn in triangle ABC, angle ADB = 120 degrees, angle B = 80 degrees. Find the angles of the DBC triangle.

The bisector divides the angle in half, so:

∠ ABD = ∠ DBC = 80 ° / 2 = 40 °;

The sum of adjacent angles is 180 °, so:

∠ ВDC = 180 ° – ∠ АDВ = 180 ° – 120 ° = 60 °;

The sum of all angles in a triangle is always 180 °, so:

∠ С = 180 ° – ∠ ВDC – ∠ DВС = 180 ° – 60 ° – 40 ° = 180 ° – 100 ° = 80 °.

Answer: 40 °, 60 °, 80 °



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