In triangle ABC, the bisector of the angles BAC and ABC intersect at point O. Find the angle ACB if the angle AOB is 125 *.

Consider the triangle AOB: the angle AOB is 125 °, the angle ABO is equal to half of the angle B (since BO is the bisector). The AO angle is equal to half of the A angle (since AO is the bisector).

The angles of a triangle add up to 180 °:

AOB + A / 2 + B / 2 = 180 °.

125 ° + A / 2 + B / 2 = 180 °.

Let us express from this the sum A / 2 + B / 2:

A / 2 + B / 2 = 180 ° – 125 °.

A / 2 + B / 2 = 55 °.

Multiply the equation by 2:

A + B = 110 °.

In triangle ABC, the sum of the angles is also 180 °:

A + B + C = 180 °.

Since A + B = 110 °, we calculate the value of the angle C:

C = 180 ° – (A + B) = 180 ° – 110 ° = 70 °.

Answer: the ACB angle is 70 °.