1. **Stating the problem:** We have a triangle where the side opposite the 30° angle is unknown, and the side opposite the 15° angle is $x$. We want to find the length of the side opposite 30°. 2. **Formula used:** We use the Law of Sines, which states: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ where $a$, $b$, and $c$ are sides opposite angles $A$, $B$, and $C$ respectively. 3. **Applying the Law of Sines:** Let the side opposite 30° be $s$. Given the side opposite 15° is $x$, we have: $$\frac{s}{\sin 30^\circ} = \frac{x}{\sin 15^\circ}$$ 4. **Solving for $s$:** $$s = x \cdot \frac{\sin 30^\circ}{\sin 15^\circ}$$ 5. **Evaluating the sines:** $$\sin 30^\circ = \frac{1}{2}$$ $$\sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}$$ 6. **Substitute values:** $$s = x \cdot \frac{\frac{1}{2}}{\frac{\sqrt{6} - \sqrt{2}}{4}} = x \cdot \frac{1}{2} \cdot \frac{4}{\sqrt{6} - \sqrt{2}} = x \cdot \frac{2}{\sqrt{6} - \sqrt{2}}$$ 7. **Rationalize the denominator:** $$s = x \cdot \frac{2}{\sqrt{6} - \sqrt{2}} \cdot \frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} + \sqrt{2}} = x \cdot \frac{2(\sqrt{6} + \sqrt{2})}{6 - 2} = x \cdot \frac{2(\sqrt{6} + \sqrt{2})}{4} = x \cdot \frac{\sqrt{6} + \sqrt{2}}{2}$$ **Final answer:** $$\boxed{s = x \cdot \frac{\sqrt{6} + \sqrt{2}}{2}}$$