A 72-degree corner C is inscribed with a circle that touches the sides of the corner at points A

A 72-degree corner C is inscribed with a circle that touches the sides of the corner at points A and B, where O is the center of the circle. Find the corner of the AOB.

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 AOBS quadrangle, the sum of the internal angles is 3600, then the AOB angle = (360 – ACB – ОАС – ОВС) = (360 – 72 – 90 – 90) = 108.

Answer: The AOB angle is 108.