An angle C of 140 ° is inscribed with a circle that touches the sides of the angle at points A

An angle C of 140 ° is inscribed with a circle that touches the sides of the angle at points A and B, point O is the center of the circle. Find the angle AOB.

1. Let’s connect points A and B. Consider the triangle ACB: CA = CB, which means the triangle is isosceles, angle ACB = 140 degrees.

2. Angle BAC = angle CBA (angles at the base of an isosceles triangle)

(180 – 140): 2 = 20 degrees

3. Consider a triangle AOB: OA = OB (radii), so the triangle is isosceles.

4. Angle ОАС = 90 degrees (angle between radius and tangent).

OAB angle = 90 – BAC angle = 90 – 20 = 70 degrees.

5. Angle AOB = 180 – (70 + 70) = 40 degrees.

Answer: the AOB angle is 40 degrees.