Through the ends of the chord AB, equal to the radius of the circle, two tangents are drawn

Through the ends of the chord AB, equal to the radius of the circle, two tangents are drawn, intersecting at point C. Find the angle ACB.

Let’s draw the radii ОА and ОВ to the edges of the chord.

Since, by condition, the length of the chord AB is equal to the radius of the circle, then the triangle ABO is equilateral, and therefore all of its internal angles are equal to 60.

Points A and B are the points of tangency of tangents AC and BC to the circle, then the radii OA and OB are perpendicular to the tangents AC and BC, and then the angle OAC = OBC = 90.

In the quadrilateral AOBC, the sum of the interior angles is 360, then the angle ACB = (360 – 90 – 90 – 60) = 120.

Answer: The ACB angle is 120.