KA and KB are tangents to the circle centered at the point O. A and B are tangency points.

| August 17, 2021 | education

KA and KB are tangents to the circle centered at the point O. A and B are tangency points. The AKB angle is 120 °, KO = 16. Find the distance between the touch points.

Let us construct the radii ОА and ОВ to the points of tangency A and B.

The radii drawn to the tangency points are perpendicular to the tangents themselves, then the triangles KOB and KOA are rectangular.

In right-angled triangles KOВ and KOA, the hypotenuse OK is common, OB = OA = R, then the triangles KOВ and KOA are equal in leg and hypotenuse, then the angle OKВ = OKA = AKВ / 2 = 120/2 = 60.

In a right-angled triangle AOK Sin60 = OA / OK.

OA = OK * Sin60 = 16 * √3 / 2 = 8 * √3 cm.

The angle AOK = 90 – 60 = 30, then the angle AOB = 30 + 30 = 60, which means that the triangle OAB is equilateral, AB = OA = 8 * √3 cm.

Answer: Between touching points 8 * √3 cm.



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