KM = KN = 16, MO = ?

The angle $$\angle KMN$$ subtends the arc $$KN$$. The angle $$\angle KNM$$ subtends the arc $$KM$$. Since $$KM = KN$$, the arcs $$KM$$ and $$KN$$ are equal. Therefore, $$\angle KMN = \angle KNM$$.
In $$\triangle KMN$$, $$KM = KN = 16$$. The angle $$\angle MKN = 120^{\circ}$$.
The sum of angles in $$\triangle KMN$$ is $$180^{\circ}$$. So, $$\angle KMN + \angle KNM + \angle MKN = 180^{\circ}$$.
$$2 \angle KMN + 120^{\circ} = 180^{\circ}$$.
$$2 \angle KMN = 60^{\circ}$$.
$$\angle KMN = 30^{\circ}$$.
The angle $$\angle MON$$ is the central angle subtending the arc $$MN$$. The angle $$\angle MKN$$ is the inscribed angle subtending the arc $$MN$$. However, $$\angle MKN$$ is $$120^{\circ}$$, which is obtuse, so it subtends the major arc $$MN$$. The reflex angle $$\angle MON$$ would be $$2 \times 120^{\circ} = 240^{\circ}$$. The angle $$\angle MON$$ is $$360^{\circ} - 240^{\circ} = 120^{\circ}$$.
Alternatively, $$\angle KMN = 30^{\circ}$$ subtends arc $$KN$$. The central angle subtending arc $$KN$$ is $$\angle KON$$. So $$\angle KON = 2 \times \angle KMN = 2 imes 30^{\circ} = 60^{\circ}$$.
Similarly, $$\angle KNM = 30^{\circ}$$ subtends arc $$KM$$. The central angle subtending arc $$KM$$ is $$\angle KOM$$. So $$\angle KOM = 2 imes \angle KNM = 2 imes 30^{\circ} = 60^{\circ}$$.
In $$\triangle KOM$$, $$KO = OM$$ (radii). $$\angle KOM = 60^{\circ}$$. Thus, $$\triangle KOM$$ is equilateral, and $$KM = KO = OM = 16$$.
Therefore, $$MO = 16$$.