Find the measure of \(\angle BAC\).

Let's analyze the given figure. We have a circle with center \(O\). The line \(AC\) is tangent to the circle at point \(A\). We also have a chord \(AB\) in the circle. The radius \(OA\) is drawn to the point of tangency \(A\). Since \(AC\) is a tangent to the circle at \(A\), and \(OA\) is a radius to the point of tangency, we know that \(\angle OAC\) is a right angle, so \(\angle OAC = 90^{\circ}\). We are also given that \(OA = AB\), meaning that triangle \(\triangle OAB\) is an isosceles triangle with \(OA = AB\). Therefore, \(\angle AOB = \angle ABO\). Let \(\angle AOB = x\). Since \(OA = AB\), \(\angle AOB = \angle ABO = x\). Then \(\angle OAB = 180^{\circ} - 2x\) because the angles in a triangle add up to 180 degrees. Since \(OA = AB\), the triangle \(\triangle OAB\) is isosceles. We are given that \(OA = AB\), this is incorrect. The problem must have \(AB\) is equal to the radius of the circle. Since segment \(AB\) is marked the same length as \(OA\), then \(OA = AB\). If \(OA = AB\), then triangle \(\triangle OAB\) is an isosceles triangle. Angle \(\angle OAB\) can be determined using angles \(\angle AOB\) and \(\angle ABO\). It is shown that angle \(\angle AOB = \angle ABO\), also angle \(\angle AOB = 45^{\circ}\). Thus angles \(\angle AOB = \angle ABO = 45^{\circ}\). So, \(\angle OAB = 180 - 45 - 45 = 90^{\circ}\). Now \(\angle BAC = \angle OAC - \angle OAB\). We have \(\angle OAC = 90^{\circ}\). Then \(\angle BAC = 90^{\circ} - \angle OAB\). Also since triangle \(\triangle OAB\) is isosceles, \(\angle OAB + \angle ABO + \angle BOA = 180\). Since angle \(\angle OBA\) appears to be 90/2 = 45 degrees, then angles \(\angle OBA = \angle AOB = 45\), hence \(\angle OAB = 90\). Therefore \(\angle BAC = 90 - 45 = 45\). So, \(\angle BAC = 45^{\circ}\). Answer: 45°