At angle C equal to 140, a circle is inscribed with center O, which touches the sides of the angle at points A

At angle C equal to 140, a circle is inscribed with center O, which touches the sides of the angle at points A and B. Find the angle AO.

From point O, the center of the circle, we draw the radii OA and OB to the points of tangency A and B.

By the property of tangents, the radius drawn to the point of tangency is perpendicular to the tangent itself, then the angle ОАС = ОВС = 90.

In the AOBС quadrangle, the sum of the internal angles is 360, then the AOB angle = (360 – ОАС – ОВС) = (360 – 140 – 90 – 90) = 40.

Answer: The AOB angle is 40.