1. **State the problem:** We have a triangle with angles 45°, 60°, and 75° (since the sum of angles in a triangle is 180°). The side opposite the 45° angle is given as $18\sqrt{2}$. We need to find the lengths $a$, $b$, and $c$ corresponding to the sides opposite the 45°, 60°, and 75° angles respectively. 2. **Use the Law of Sines:** The Law of Sines states that for any triangle, $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ where $a$, $b$, and $c$ are sides opposite angles $A$, $B$, and $C$ respectively. 3. **Assign sides and angles:** - Side opposite 45° is $a = 18\sqrt{2}$ - Side opposite 60° is $b$ - Side opposite 75° is $c$ 4. **Calculate $b$ using Law of Sines:** $$\frac{a}{\sin 45^\circ} = \frac{b}{\sin 60^\circ}$$ $$b = a \times \frac{\sin 60^\circ}{\sin 45^\circ}$$ Substitute $a = 18\sqrt{2}$: $$b = 18\sqrt{2} \times \frac{\sin 60^\circ}{\sin 45^\circ}$$ Recall $\sin 45^\circ = \frac{\sqrt{2}}{2}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$: $$b = 18\sqrt{2} \times \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}} = 18\sqrt{2} \times \frac{\sqrt{3}}{\sqrt{2}}$$ Simplify: $$b = 18 \times \cancel{\sqrt{2}} \times \frac{\sqrt{3}}{\cancel{\sqrt{2}}} = 18\sqrt{3}$$ 5. **Calculate $c$ using Law of Sines:** $$\frac{a}{\sin 45^\circ} = \frac{c}{\sin 75^\circ}$$ $$c = a \times \frac{\sin 75^\circ}{\sin 45^\circ}$$ Substitute $a = 18\sqrt{2}$: $$c = 18\sqrt{2} \times \frac{\sin 75^\circ}{\sin 45^\circ}$$ Recall $\sin 75^\circ = \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$: $$c = 18\sqrt{2} \times \frac{\frac{\sqrt{6} + \sqrt{2}}{4}}{\frac{\sqrt{2}}{2}} = 18\sqrt{2} \times \frac{\sqrt{6} + \sqrt{2}}{4} \times \frac{2}{\sqrt{2}}$$ Simplify: $$c = 18\sqrt{2} \times \frac{2}{4} \times \frac{\sqrt{6} + \sqrt{2}}{\sqrt{2}} = 18\sqrt{2} \times \frac{1}{2} \times \frac{\sqrt{6} + \sqrt{2}}{\sqrt{2}}$$ $$c = 9\sqrt{2} \times \frac{\sqrt{6} + \sqrt{2}}{\sqrt{2}} = 9 \times \cancel{\sqrt{2}} \times \frac{\sqrt{6} + \sqrt{2}}{\cancel{\sqrt{2}}} = 9(\sqrt{6} + \sqrt{2})$$ **Final answers:** $$a = 18\sqrt{2}$$ $$b = 18\sqrt{3}$$ $$c = 9(\sqrt{6} + \sqrt{2})$$