1. **Problem statement:** Given a circle with center O and points A, B, C on the circumference, we know $\angle ABO = 32^\circ$ and $\angle BOC = 160^\circ$. We need to find $\angle ACO$. 2. **Key properties and formulas:** - $OA, OB, OC$ are radii of the circle, so $OA = OB = OC$. - Triangle $OBC$ is isosceles with $OB = OC$. - The central angle $\angle BOC = 160^\circ$ subtends arc $BC$. - The angle at the circumference subtending the same arc $BC$ is half the central angle, so $\angle BAC = \frac{1}{2} \times 160^\circ = 80^\circ$. 3. **Find $\angle OBC$:** Since $\angle ABO = 32^\circ$ and $OA = OB$, triangle $OAB$ is isosceles with $OA = OB$. Thus, $\angle OAB = \angle ABO = 32^\circ$. 4. **Find $\angle AOB$:** Sum of angles in triangle $OAB$ is $180^\circ$: $$\angle AOB + 32^\circ + 32^\circ = 180^\circ$$ $$\angle AOB = 180^\circ - 64^\circ = 116^\circ$$ 5. **Find arc $AB$:** Central angle $\angle AOB = 116^\circ$ subtends arc $AB$ of $116^\circ$. 6. **Find arc $AC$:** Total circle is $360^\circ$, so arc $AC = 360^\circ - (arc AB + arc BC)$ $$= 360^\circ - (116^\circ + 160^\circ) = 360^\circ - 276^\circ = 84^\circ$$ 7. **Find $\angle ACO$:** Triangle $OAC$ is isosceles with $OA = OC$. Central angle $\angle AOC = $ arc $AC = 84^\circ$. Sum of angles in $\triangle OAC$: $$\angle ACO + \angle OAC + \angle AOC = 180^\circ$$ Since $\angle ACO = \angle OAC$, let each be $x$: $$x + x + 84^\circ = 180^\circ$$ $$2x = 180^\circ - 84^\circ = 96^\circ$$ $$x = 48^\circ$$ **Final answer:** $\angle ACO = 48^\circ$ which corresponds to option C.