1. **State the problem:** We have a circle with points A, B, C, D on the circumference and a tangent EBF at point B. Given angles are $\angle BAC = 40^\circ$ and the angle between tangent EBF and chord BC at B is $66^\circ$. We need to find $\angle DCB$. 2. **Recall the tangent-chord angle theorem:** The angle between a tangent and a chord through the point of contact equals the angle in the alternate segment of the circle. So, $$\angle EBF = \angle BCA = 66^\circ$$ 3. **Use the fact that angles subtended by the same chord are equal:** Since $\angle BAC = 40^\circ$, the angle subtended by chord AC at point D on the circumference is also $40^\circ$: $$\angle BDC = 40^\circ$$ 4. **Use the fact that quadrilateral ABCD is cyclic:** The opposite angles of a cyclic quadrilateral sum to $180^\circ$. So, $$\angle BDC + \angle BCA = 180^\circ$$ 5. **Calculate $\angle DCB$:** Since $\angle BCA = 66^\circ$ and $\angle BDC = 40^\circ$, the angle $\angle DCB$ is the angle at C between points D and B. Using the triangle $BCD$, the sum of angles is $180^\circ$: $$\angle DCB = 180^\circ - \angle BCA - \angle BDC = 180^\circ - 66^\circ - 40^\circ = 74^\circ$$ **Final answer:** $$\boxed{74^\circ}$$