4. Find \(\angle\) AMB.

Solution:

1. OA and OB are radii. AM and BM are tangents to the circle from point M.
2. Since AM and BM are tangents from M, OM bisects \(\angle AMB\) and \(\angle AOB\). Also, OM is perpendicular to AB if M is on the perpendicular bisector of AB.
3. In this diagram, it looks like OA = OB (radii). AM and BM are tangents from M. Therefore, \(\angle OAM = 90^{\circ}\) and \(\angle OBM = 90^{\circ}\).
4. In quadrilateral OAMB, the sum of angles is 360 degrees. \(\angle AOB + \angle OAM + \angle AMB + \angle OBM = 360^{\circ}\).
5. \(\angle AOB + 90^{\circ} + \angle AMB + 90^{\circ} = 360^{\circ}\)
6. \(\angle AOB + \angle AMB = 180^{\circ}\).
7. We need to find \(\angle AOB\). From the diagram in problem 3, it was indicated that OA=AB=radius, which made \(\triangle OAB\) equilateral and \(\angle AOB = 60^{\circ}\). If we assume this is a continuation or similar scenario, then \(\angle AOB = 60^{\circ}\).
8. If \(\angle AOB = 60^{\circ}\), then \(60^{\circ} + \angle AMB = 180^{\circ}\).
9. \(\angle AMB = 180^{\circ} - 60^{\circ} = 120^{\circ}\).
10. Let's verify this assumption. The diagram shows OA = radius, and tick marks on AB suggest AB = radius. So \(\triangle OAB\) is equilateral. \(\angle AOB = 60^{\circ}\).
11. In \(\triangle OAM\), OA is radius. \(\angle OAM = 90^{\circ}\). \(\angle AOM = \frac{1}{2} \angle AOB = \frac{1}{2} \times 60^{\circ} = 30^{\circ}\).
12. In \(\triangle OAM\), \(\tan(\angle AOM) = \frac{AM}{OA}\) => \(\tan(30^{\circ}) = \frac{AM}{OA}\) => \( \frac{1}{\sqrt{3}} = \frac{AM}{OA}\) => \( AM = \frac{OA}{\sqrt{3}} \).
13. In \(\triangle OAM\), \(\angle AMO = 180^{\circ} - 90^{\circ} - 30^{\circ} = 60^{\circ}\).
14. Since OM bisects \(\angle AMB\), \(\angle AMB = 2 \times \angle AMO = 2 \times 60^{\circ} = 120^{\circ}\).

Answer: \(\angle AMB = 120^{\circ}\).