In the inner region of triangle ABC, point O is taken, equidistant from its sides, Find the angle AOC

In the inner region of triangle ABC, point O is taken, equidistant from its sides, Find the angle AOC if the angle ABO = 39 degrees

Since point O is equidistant from the sides of the triangle, the point is the center of the inscribed circle, which coincides with the intersection point of the bisectors of the triangle.

Then the angle ABC = ABO * 2 = 39 * 2 = 78.

Let us determine the sum of the angles BAC and BCA.

Angle (BAC + BCA) = (180 – 78) = 102.

Since AO and CO are the bisectors of the angles, the sum of the angles OAC and OCA is equal to half the sum of the angles BAC and BCA.

Angle (OСA + OСA) = 102/2 = 51.

Then the angle AOC = 180 – 51 = 129.

Answer: The AOC angle is 129.