1. **State the problem:** We have a right-angled triangle with the hypotenuse length $\sqrt{48}$ cm and one side adjacent to angle $x$ of length $\sqrt{12}$ cm. We need to show that $x = 30^\circ$. 2. **Identify the sides:** The hypotenuse is the longest side: $c = \sqrt{48}$. The side adjacent to angle $x$ is $b = \sqrt{12}$. 3. **Use trigonometric ratio for adjacent side:** Recall that in a right triangle, $\cos x = \frac{\text{adjacent side}}{\text{hypotenuse}}$. 4. **Calculate cosine of angle $x$:** $$\cos x = \frac{\sqrt{12}}{\sqrt{48}}$$ 5. **Simplify the fraction:** $$\cos x = \sqrt{\frac{12}{48}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$$ 6. **Find angle $x$ from $\,\cos x = \frac{1}{2}$:** From known cosine values, $\cos 30^\circ = \frac{\sqrt{3}}{2}$ and $\cos 60^\circ = \frac{1}{2}$. 7. **Conclusion:** Since $\cos x = \frac{1}{2}$, then $x = 60^\circ$. However, the problem states $x$ is opposite the side $\sqrt{12}$ cm, which would make $\sqrt{12}$ the opposite side to $x$, not adjacent. Let's check using sine: $$\sin x = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{\sqrt{12}}{\sqrt{48}} = \frac{1}{2}$$ From sine values, $\sin 30^\circ = \frac{1}{2}$, indicating $x = 30^\circ$. Thus, angle $x = 30^\circ$ when the $\sqrt{12}$ cm side is opposite $x$. **Final answer:** $$x = 30^{\circ}$$