Question: In the figure below not drawn to scale, $\angle BOC = 66^\circ$ and $\angle COD$ is 3 times as big as $\angle AOB$. Find $\angle COD$.

1. **State the problem:** We are given a figure with angles at point $O$. We know $\angle BOC = 66^\circ$ and $\angle COD$ is 3 times $\angle AOB$. We need to find $\angle COD$. 2. **Identify the angles around point $O$:** The rays $OA$, $OB$, $OC$, and $OD$ form angles around point $O$. Since $OA$ is vertical upward and $OD$ is horizontal left, $\angle AOD$ is a right angle, so: $$\angle AOB + \angle BOC + \angle COD + \angle DOA = 360^\circ$$ But since $OA$ and $OD$ are perpendicular, $\angle AOD = 90^\circ$. 3. **Express known angles:** - $\angle BOC = 66^\circ$ - $\angle COD = 3 \times \angle AOB$ - $\angle DOA = 90^\circ$ 4. **Set variables:** Let $x = \angle AOB$. Then $\angle COD = 3x$. 5. **Sum of angles around $O$:** $$x + 66 + 3x + 90 = 360$$ 6. **Simplify the equation:** $$4x + 156 = 360$$ 7. **Isolate $x$:** $$4x = 360 - 156$$ $$4x = 204$$ 8. **Divide both sides by 4:** $$x = \frac{204}{4}$$ $$x = 51$$ 9. **Find $\angle COD$:** $$\angle COD = 3x = 3 \times 51 = 153^\circ$$ **Final answer:** $$\boxed{153^\circ}$$