2. OM = 18. Find \(\angle\) NMK.

Solution:

1. OM = 18 is given. The radius of the circle is ON = OK = 9.
2. Since ON and OK are radii and NM and MK are tangents from point M to the circle, ON is perpendicular to NM, and OK is perpendicular to MK. So, \(\angle ONM = 90^{\circ}\) and \(\angle OKM = 90^{\circ}\).
3. Triangle ONM and triangle OKM are right-angled triangles.
4. In \(\triangle ONM\), we have ON = 9 and OM = 18. We can find NM using the Pythagorean theorem: \( NM^2 + ON^2 = OM^2 \) => \( NM^2 + 9^2 = 18^2 \) => \( NM^2 + 81 = 324 \) => \( NM^2 = 243 \) => \( NM = \sqrt{243} = \sqrt{81 \times 3} = 9\sqrt{3} \).
5. Similarly, in \(\triangle OKM\), NK = 9 and OM = 18, so MK = $$9\sqrt{3}$$.
6. Also, tangents from an external point to a circle are equal in length, so NM = MK.
7. In \(\triangle ONM\), we can find \(\angle NOM\) using sine: \( \sin(\angle NOM) = \frac{NM}{OM} = \frac{9\sqrt{3}}{18} = \frac{\sqrt{3}}{2} \). Therefore, \(\angle NOM = 60^{\circ}\).
8. Similarly, \(\angle KOM = 60^{\circ}\).
9. Thus, \(\angle NOK = \angle NOM + \angle KOM = 60^{\circ} + 60^{\circ} = 120^{\circ}\).
10. In quadrilateral ONMK, the sum of angles is 360 degrees. \(\angle ONM = 90^{\circ}\), \(\angle OKM = 90^{\circ}\), \(\angle NOK = 120^{\circ}\).
11. So, \(\angle NMK = 360^{\circ} - 90^{\circ} - 90^{\circ} - 120^{\circ} = 60^{\circ}\).
12. Alternatively, consider \(\triangle ONM\) and \(\triangle OKM\). They are congruent by RHS (Right angle, Hypotenuse, Side) since ON = OK (radii), OM is common, and \(\angle ONM = \angle OKM = 90^{\circ}\).
13. Therefore, \(\angle NMO = \angle KMO\).
14. In \(\triangle ONM\), \(\sin(\angle NMO) = \frac{ON}{OM} = \frac{9}{18} = \frac{1}{2}\). Thus, \(\angle NMO = 30^{\circ}\).
15. Since \(\angle NMK = \angle NMO + \angle KMO\) and \(\angle NMO = \angle KMO\), then \(\angle NMK = 30^{\circ} + 30^{\circ} = 60^{\circ}\).

Answer: \(\angle NMK = 60^{\circ}\).