In a triangle ABC m (angle A) = 57. Find the obtuse angle between the bisectors of angles B and C.

Since the sum of the inner angles of the triangle is 180, then the sum of the angles (ABC + ACB) = (180 – BAC) = (180 – 57) = 123.

Since the segments CH and BM are the bisectors of the angles, the sum of the angles (BCН + CBM) = (ABC + ACB) / 2 = 123/2 = 61.5.

Then in the triangle BОС the angle BОС = (180 – (ВСН + СВМ)) = (180 – 61.5) = 118.5.

Answer: The obtuse angle between the bisectors of the angles is 118.5.



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