1. **State the problem:** We have a right triangle with a 60° angle, the side opposite this angle is $6\sqrt{3}$, the adjacent side is $y$, and the hypotenuse is $x$. We need to find $x$ and $y$. 2. **Recall trigonometric ratios:** For an angle $\theta$ in a right triangle, - Opposite side = $\text{hypotenuse} \times \sin(\theta)$ - Adjacent side = $\text{hypotenuse} \times \cos(\theta)$ 3. **Apply the sine ratio:** $$\sin(60^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{6\sqrt{3}}{x}$$ We know $\sin(60^\circ) = \frac{\sqrt{3}}{2}$, so $$\frac{\sqrt{3}}{2} = \frac{6\sqrt{3}}{x}$$ 4. **Solve for $x$:** Multiply both sides by $x$: $$x \times \frac{\sqrt{3}}{2} = 6\sqrt{3}$$ Divide both sides by $\frac{\sqrt{3}}{2}$: $$x = \frac{6\sqrt{3}}{\frac{\sqrt{3}}{2}}$$ Simplify the denominator: $$x = 6\sqrt{3} \times \frac{2}{\sqrt{3}}$$ Cancel $\sqrt{3}$: $$x = 6 \times 2 = 12$$ 5. **Apply the cosine ratio to find $y$:** $$\cos(60^\circ) = \frac{y}{x}$$ We know $\cos(60^\circ) = \frac{1}{2}$, so $$\frac{1}{2} = \frac{y}{12}$$ Multiply both sides by 12: $$y = 12 \times \frac{1}{2} = 6$$ 6. **Final answer:** $$x = 12, \quad y = 6$$ This corresponds to option C.