3. Find \(\angle\) BAC.

Solution:

1. OA and OB are radii. AB is a chord. AC is a tangent to the circle at point A.
2. Since AC is a tangent and OA is the radius to the point of contact, \(\angle OAC = 90^{\circ}\).
3. In \(\triangle OAB\), OA = OB (radii), so it is an isosceles triangle.
4. Let's assume \(\angle OAB = \angle OBA = x\). Then \(\angle AOB = 180^{\circ} - 2x\).
5. We are given that \(\angle BAC\) is to be found. \(\angle BAC = \angle OAC - \angle OAB = 90^{\circ} - x\).
6. We need more information to solve this. Let's check the diagram for any implicit information. There are tick marks on OA and AB, which implies OA = AB.
7. Since OA = AB, \(\triangle OAB\) is an isosceles triangle with OA = AB. Since OA is a radius, AB is also equal to the radius.
8. In \(\triangle OAB\), OA = OB = AB (all are radii). This means \(\triangle OAB\) is an equilateral triangle.
9. Therefore, \(\angle AOB = 60^{\circ}\) and \(\angle OAB = \angle OBA = 60^{\circ}\).
10. Now, \(\angle BAC = \angle OAC - \angle OAB\).
11. Since AC is a tangent at A, \(\angle OAC = 90^{\circ}\).
12. So, \(\angle BAC = 90^{\circ} - 60^{\circ} = 30^{\circ}\).

Answer: \(\angle BAC = 30^{\circ}\).