1. **State the problem:** We have a circle with center $O$ and points $A$, $B$, and $C$ on the circumference. Given angles: - $\angle BAO = 36^\circ$ - $\angle BCO = 21^\circ$ We need to find the size of $\angle ACO$. 2. **Understand the setup:** Since $O$ is the center and $A$, $B$, $C$ lie on the circle, segments $OA$, $OB$, and $OC$ are radii and thus equal in length. 3. **Analyze triangle $OAB$:** Since $OA = OB$, triangle $OAB$ is isosceles with $OA = OB$. Given $\angle BAO = 36^\circ$, then $\angle ABO = 36^\circ$ (base angles of isosceles triangle). Calculate $\angle AOB$: $$\angle AOB = 180^\circ - 2 \times 36^\circ = 180^\circ - 72^\circ = 108^\circ$$ 4. **Analyze triangle $OBC$:** Similarly, $OB = OC$, so triangle $OBC$ is isosceles. Given $\angle BCO = 21^\circ$, then $\angle BOC = 21^\circ$ (since base angles are equal). Calculate $\angle OBC$: $$\angle OBC = 180^\circ - 2 \times 21^\circ = 180^\circ - 42^\circ = 138^\circ$$ 5. **Find $\angle ACB$:** Since $A$, $B$, $C$ lie on the circle, $\angle ACB$ is an inscribed angle subtending arc $AB$. Central angle $\angle AOB = 108^\circ$ subtends the same arc $AB$. By the circle theorem, the inscribed angle is half the central angle: $$\angle ACB = \frac{1}{2} \times 108^\circ = 54^\circ$$ 6. **Find $\angle ACO$:** In triangle $OBC$, angles are $\angle BCO$, $\angle OBC$, and $\angle BOC$. We know $\angle BCO = 21^\circ$ and $\angle OBC = 138^\circ$. Since $\angle ACB = 54^\circ$ and $\angle BCO = 21^\circ$, then: $$\angle ACO = \angle ACB - \angle BCO = 54^\circ - 21^\circ = 33^\circ$$ **Final answer:** $$\boxed{33^\circ}$$