1. The problem involves finding the values of $x$ and $y$ in a right triangle with angles 45° and 60°, and given side lengths. 2. We use trigonometric ratios and the Pythagorean theorem to find unknown sides. 3. For the triangle with a 45° angle and side $y=30$, since it's a 45°-45°-90° triangle, the sides are in ratio $1:1:\sqrt{2}$. 4. Therefore, $x = y = 30$ because the legs opposite 45° angles are equal. 5. For the triangle with angles 45° and 60°, and side $z=32$, use the Law of Sines: $$\frac{z}{\sin 45^\circ} = \frac{x}{\sin 60^\circ}$$ 6. Substitute values: $$\frac{32}{\sin 45^\circ} = \frac{x}{\sin 60^\circ}$$ 7. Calculate sines: $$\sin 45^\circ = \frac{\sqrt{2}}{2}, \quad \sin 60^\circ = \frac{\sqrt{3}}{2}$$ 8. So: $$\frac{32}{\frac{\sqrt{2}}{2}} = \frac{x}{\frac{\sqrt{3}}{2}}$$ 9. Simplify left side: $$32 \times \frac{2}{\sqrt{2}} = \frac{64}{\sqrt{2}}$$ 10. Cross multiply: $$x = \frac{64}{\sqrt{2}} \times \frac{\sqrt{3}}{2} = \frac{64 \sqrt{3}}{2 \sqrt{2}}$$ 11. Simplify fraction: $$x = 32 \times \frac{\sqrt{3}}{\sqrt{2}} = 32 \sqrt{\frac{3}{2}}$$ 12. Rationalize denominator: $$x = 32 \times \frac{\sqrt{6}}{2} = 16 \sqrt{6}$$ 13. Therefore, $x = 16 \sqrt{6}$. 14. To find $y$, use the Pythagorean theorem or Law of Sines similarly if needed, but problem only asks for $x$ and $y$. Final answers: $$x = 30$$ $$y = 30$$ $$x = 16 \sqrt{6}$$ (for the second triangle)