Through point A to the circle whose center is point O, a tangent point B is drawn, the tangent point

Through point A to the circle whose center is point O, a tangent point B is drawn, the tangent point of the segment oa intersects the circle at point F, angle AFB 120 degrees, find the area of triangle AFB.

The area of ​​triangle AFB can be found by subtracting the area of ​​triangle OBF from the area of ​​triangle AOB.

AFB = AOB – OBF

If point B is tangent, then OBA = 90 °.

So the area of ​​a right-angled triangle is OBA = AOB = 1/2 * OB * BA.

The OBF triangle has two vertices, OB and OF, on a circle. Since the circle has the same distance from any point to the center, the sides of the OBF triangle, namely OB and OF, are also equal. This means that the OBF triangle is isosceles.

From the problem statement, we know that the angle AFB = 120 °, which means the adjacent angle BFO = 180 – 120 ° = 60 °.

In an isosceles triangle, the angles BFO and FBO must be equal, hence the angle FBO is also 60 °.

The angles of the triangle must add up to 180 degrees.
This means that the angle of an isosceles triangle BOF = 180 – angle BFO – angle FBO = 180 – 60 – 60 = 60 °.
A triangle in which all 3 angles are equal is called equilateral.
Therefore, all sides of the triangle BOF are equal to each other, that is, BO = OF = FB.

Area triangle BOF = 1/2 * sin (60 °) * BO * OF = 1/2 * √3 / 2 * BO ^ 2

Desired area of ​​a triangle AFB = AOB – OBF

AFB = 1/2 * OB * BA – 1/2 * √3 / 2 * BO ^ 2

Also consider that in a right-angled triangle AOB we know the angle AOB = 60 °.
So tangent tan (60 °) = AB / BO = √3
AB = √3 * BO

Area AFB = 1/2 * BO * √3 * BO – 1/2 * √3 / 2 * BO ^ 2 = 1/2 * BO ^ 2 * (√3 – √3 / 2) = 1/2 * √ 3/2 * BO ^ 2 = √3 / 4 * BO ^ 2 = √3 / 4 * OF ^ 2 = √3 / 4 * FB ^ 2

Answer: area of ​​triangle AFB = √3 / 4 * FB ^ 2