A circle with a center O of radius 12 cm is described near a triangle МNК so that the angle МN = 120

A circle with a center O of radius 12 cm is described near a triangle МNК so that the angle МN = 120 degrees, the angle NОК = 90 degrees. Find the lengths of the sides МN and NК of the triangle.

Since, by condition, the central angle HОК = 90, then the triangle HОК is rectangular and isosceles, since OK = ОН = 12 cm.

Then HK ^ 2 = OK ^ 2 + OH ^ 2 = 144 + 144 = 288.

HK = 12 * √2 cm.

The MOH triangle is isosceles with an apex angle of 120, then, according to the cosine theorem: MH ^ 2 = OH ^ 2 + OM ^ 2 – 2 * OH * OM * Cos120 = 144 + 144 – 2 * 12 * 12 * (1/2) = 432.

MH = 12 * √3 cm.

Answer: The length of the side HK is 12 * √2 cm, MH is 12 * √3 cm.