In the figure, O is the center of the circle, and M is a point on the chord AC. OM is perpendicular to AC. If AC = 10 and the angle ABC = 60 degrees, find the length of OM.

Solution:

In a circle, the angle subtended by an arc at the center is twice the angle subtended by the same arc at any point on the remaining part of the circle. The angle subtended by arc AC at point B is \( \angle ABC = 60^ \). Therefore, the angle subtended by arc AC at the center O is \( \angle AOC = 2 \angle ABC = 2 60^ = 120^ \).

Since OM is perpendicular to AC, M is the midpoint of AC. Therefore, \( AM = MC = \frac{AC}{2} = \frac{10}{2} = 5 \).

In the right-angled triangle OMC, we have \( \angle OMC = 90^ \), \( MC = 5 \), and \( \angle MOC = \frac{\angle AOC}{2} = \frac{120^}{2} = 60^ \) (because OM bisects \( \angle AOC \) in an isosceles triangle AOC).

Using trigonometry in \( \triangle OMC \):

\( \tan(\angle MOC) = \frac{MC}{OM} \)

\( \tan(60^) = \frac{5}{OM} \)

\( \sqrt{3} = \frac{5}{OM} \)

\( OM = \frac{5}{\sqrt{3}} = \frac{5\sqrt{3}}{3} \)

Alternatively, using \( \angle OCM \):

In \( \triangle AOC \), OA = OC (radii), so it's an isosceles triangle. \( \angle OAC = \angle OCA = \frac{180^ - 120^}{2} = \frac{60^}{2} = 30^ \).

In the right-angled triangle OMC:

\( \tan(\angle OCM) = \frac{OM}{MC} \)

\( \tan(30^) = \frac{OM}{5} \)

\( \frac{1}{\sqrt{3}} = \frac{OM}{5} \)

\( OM = \frac{5}{\sqrt{3}} = \frac{5\sqrt{3}}{3} \)

Ответ: \( OM = \frac{5\sqrt{3}}{3} \).