1. **Problem:** Given a circle with center O, angles \(\angle OAB = 30^\circ\) and \(\angle OCB = 40^\circ\), find \(\angle AOC\). 2. **Formula and rules:** In a circle, the angle subtended by an arc at the center is twice the angle subtended at the circumference on the same side. Also, triangle angle sum is \(180^\circ\). 3. **Step 1:** Join \(OB\). Since \(O\) is the center, \(OA = OB = OC\) (radii of the circle). Triangles \(OAB\) and \(OCB\) are isosceles. 4. **Step 2:** In \(\triangle OAB\), angles at \(A\) and \(B\) are equal except \(\angle OAB = 30^\circ\). Let \(\angle OBA = x\). Then \(\angle AOB = 180^\circ - 30^\circ - x = 150^\circ - x\). But since \(OA = OB\), \(\angle OAB = \angle OBA = 30^\circ\), so \(x = 30^\circ\). Thus, \(\angle AOB = 180^\circ - 30^\circ - 30^\circ = 120^\circ\). 5. **Step 3:** In \(\triangle OCB\), similarly, \(\angle OCB = 40^\circ\), so \(\angle OBC = 40^\circ\) (isosceles). Then \(\angle BOC = 180^\circ - 40^\circ - 40^\circ = 100^\circ\). 6. **Step 4:** Now, \(\angle AOC = \angle AOB + \angle BOC = 120^\circ + 100^\circ = 220^\circ\). But \(\angle AOC\) is the smaller angle inside the circle, so \(\angle AOC = 360^\circ - 220^\circ = 140^\circ\). **Final answer:** \(\boxed{140^\circ}\)