In an acute-angled triangle ABC, the mid-perpendiculars of the sides BC and AC

In an acute-angled triangle ABC, the mid-perpendiculars of the sides BC and AC intersect at point O. Find the side OC if AB = 10 and the angle BOA = 120 °.

Consider an acute-angled triangle ABC.

The mid-perpendiculars to the sides BC and AC meet at point O.

By the property of the median perpendicular we have:

OB = OС and OA = OС. Hence,

OA = OB = OС. It follows from this that O is the center of a circle described around the triangle ABC and its radius R = OA = OV = OS.

Note that the angle of the ВCA is based on the arc AB, the central angle of which is BOA = 120 °. Hence:

BCA = 1/2 * BOA = 1/2 * 120 ° = 60 °.

By the sine theorem we have:

AB / sin (BCA) = 2 * R,

10 / sin (60 °) = 2 * R,

10 / (√3 / 2) = 2 * R,

10 / √3 = R,

OC = R = 10 * √3 / 3.