1. **Problem statement:** Given $mCG = 6x + 5$, $mAR = 4x + 15$, and $m\angle AEC = 120$, find: a) $x$ b) $mCG$ c) $mAR$ 2. **Relevant formula and rules:** Since $E$ is the intersection point inside the circle, the measure of angle $AEC$ formed by two chords is half the sum of the measures of the arcs intercepted by the angle and its vertical angle. This means: $$m\angle AEC = \frac{1}{2}(mCG + mAR)$$ 3. **Set up the equation:** Substitute the given expressions: $$120 = \frac{1}{2}((6x + 5) + (4x + 15))$$ 4. **Simplify inside the parentheses:** $$120 = \frac{1}{2}(6x + 5 + 4x + 15)$$ $$120 = \frac{1}{2}(10x + 20)$$ 5. **Multiply both sides by 2 to eliminate the fraction:** $$2 \times 120 = \cancel{2} \times \frac{1}{\cancel{2}}(10x + 20)$$ $$240 = 10x + 20$$ 6. **Solve for $x$:** Subtract 20 from both sides: $$240 - 20 = 10x + \cancel{20} - \cancel{20}$$ $$220 = 10x$$ Divide both sides by 10: $$\frac{220}{\cancel{10}} = \frac{10x}{\cancel{10}}$$ $$22 = x$$ 7. **Find $mCG$:** Substitute $x = 22$ into $mCG = 6x + 5$: $$mCG = 6(22) + 5 = 132 + 5 = 137$$ 8. **Find $mAR$:** Substitute $x = 22$ into $mAR = 4x + 15$: $$mAR = 4(22) + 15 = 88 + 15 = 103$$ **Final answers:** - a) $x = 22$ - b) $mCG = 137$ - c) $mAR = 103$