1. **Problem statement:** We have a circle with center $O$. Lines $BC$ and $DE$ are tangents to the circle at points $B$ and $D$ respectively. We know $\angle AOB = 52^\circ$ and $\angle EDB = 27^\circ$. We need to find the size of angle $\theta$ at point $C$, where tangents $BC$ and $DE$ meet. 2. **Key properties and formulas:** - The angle at the center $O$ subtended by an arc is twice the angle at the circumference subtended by the same arc. So, $\angle AOB = 2 \times \angle ADB$. - Tangent lines to a circle are perpendicular to the radius at the point of tangency. So, $\angle OBD = 90^\circ$ and $\angle ODB = 90^\circ$. - The angle between two tangents from an external point equals $180^\circ$ minus the angle subtended by the line segment joining the points of tangency at the center. 3. **Find $\angle ADB$:** $$\angle ADB = \frac{1}{2} \times \angle AOB = \frac{1}{2} \times 52^\circ = 26^\circ$$ 4. **Analyze triangle $BDE$:** Given $\angle EDB = 27^\circ$ and $\angle ADB = 26^\circ$, note that $D$ lies on the circle and $B$ and $E$ are points on tangents. 5. **Find $\angle BDE$:** Since $DE$ is tangent at $D$, $\angle BDE$ is the angle between tangent $DE$ and chord $DB$. The angle between tangent and chord equals the angle in the alternate segment, so: $$\angle BDE = \angle ADB = 26^\circ$$ 6. **Find $\angle BEC$:** Since $BC$ and $DE$ are tangents meeting at $C$, $\theta = \angle BCE$ is the angle between the two tangents. 7. **Use the tangent-tangent angle theorem:** The angle between two tangents from an external point equals $180^\circ$ minus the angle subtended by the chord joining the points of tangency at the center: $$\theta = 180^\circ - \angle BOD$$ 8. **Find $\angle BOD$:** $\angle BOD$ is the central angle subtended by arc $BD$. Since $\angle AOB = 52^\circ$ and $\angle BOD$ includes $\angle AOB$ plus $\angle AOD$, but $A$, $B$, and $D$ lie on the circle, and $\angle BOD = 2 \times \angle BCD$. However, from the given data, $\angle BOD = 2 \times (\angle BCD)$, but $\angle BCD = \theta$. 9. **Calculate $\theta$:** Using the tangent-tangent angle theorem: $$\theta = 180^\circ - \angle BOD$$ But $\angle BOD = 2 \times \angle BCD = 2 \times \theta$, so: $$\theta = 180^\circ - 2\theta$$ Add $2\theta$ to both sides: $$\theta + 2\theta = 180^\circ$$ $$3\theta = 180^\circ$$ Divide both sides by 3: $$\theta = \frac{180^\circ}{3} = 60^\circ$$ 10. **Final answer:** $$\boxed{\theta = 60^\circ}$$ **Reason:** The angle between two tangents from an external point equals $180^\circ$ minus the central angle between the points of tangency.