1. **Problem statement:** Given a right triangle with hypotenuse $6\sqrt{3}$ and an angle of $30^\circ$, find the lengths of sides $x$ (adjacent to $30^\circ$) and $y$ (opposite to $30^\circ$). 2. **Formula and rules:** In a right triangle, for an angle $\theta$, the side opposite is $\text{hypotenuse} \times \sin(\theta)$ and the side adjacent is $\text{hypotenuse} \times \cos(\theta)$. 3. **Calculate $x$ (adjacent side):** $$x = 6\sqrt{3} \times \cos(30^\circ)$$ Recall $\cos(30^\circ) = \frac{\sqrt{3}}{2}$. $$x = 6\sqrt{3} \times \frac{\sqrt{3}}{2}$$ Simplify: $$x = 6 \times \cancel{\sqrt{3}} \times \frac{\cancel{\sqrt{3}}}{2} = 6 \times \frac{3}{2} = 9$$ 4. **Calculate $y$ (opposite side):** $$y = 6\sqrt{3} \times \sin(30^\circ)$$ Recall $\sin(30^\circ) = \frac{1}{2}$. $$y = 6\sqrt{3} \times \frac{1}{2} = 3\sqrt{3}$$ 5. **Answer:** $$x = 9, \quad y = 3\sqrt{3}$$ 6. **Match with options:** Option C) $x=9$, $y=3\sqrt{3}$ is correct.