In triangle ABC, angle C is 76, AL and BM are the bisectors of angles A and B, intersecting at point O. Find angle AOB.

The sum of the interior angles of a triangle is 180.

Then (BCA + BAC + ABC) = 180.

(BAC + ABC) = (180 – BCA) = (180 – 76) = 104.

Since the Segments AL and VM are the bisectors of the angles ABC and BAC, then the angle OBA = ABC / 2, the angle OAB = BAC / 2.

Angle (OBA + OAB) = ABC / 2 + BAC / 2 = (ABC + BAC) / 2 = 104/2 = 52.

Then in the triangle AOB the angle AOB = (180 – 52) = 128.

Answer: The value of the angle AOB is equal to 128.