1. **State the problem:** We have a circle with center $O$, a tangent line $ABC$ touching the circle at $B$, and chords $AD$, $BE$, and $CE$. We need to find the size of angle $\theta$ at the center $O$ formed by lines $OD$ and $OE$. Given angles are $\angle C = 57^\circ$ and $\angle A = 48^\circ$. 2. **Recall the tangent-secant 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. That is, $\angle ABC = \angle ADB$. Since $ABC$ is tangent at $B$, $\angle ABC$ is the angle between tangent and chord $BC$. 3. **Identify angles:** Given $\angle A = 48^\circ$ and $\angle C = 57^\circ$, and $ABC$ is tangent at $B$, then $\angle ABC$ is the angle between tangent and chord $BC$. By the alternate segment theorem, $\angle ABC = \angle BDE$. 4. **Calculate $\angle ABC$:** Since $ABC$ is a straight line, $\angle ABC = 180^\circ - (\angle A + \angle C) = 180^\circ - (48^\circ + 57^\circ) = 180^\circ - 105^\circ = 75^\circ$. 5. **Use the alternate segment theorem:** $\angle BDE = 75^\circ$. 6. **Use the fact that $\theta$ is the central angle subtending arc $DE$:** The central angle $\theta = 2 \times$ the inscribed angle subtending the same arc. Since $\angle BDE$ is an inscribed angle subtending arc $DE$, $$\theta = 2 \times 75^\circ = 150^\circ.$$ 7. **Final answer:** $\boxed{150^\circ}$. **Reason:** The tangent-secant angle theorem and the property that the central angle is twice the inscribed angle subtending the same arc.