1. **Problem statement:** Find the values of $x$ and $y$ in the given triangles. 2. **First triangle:** Right triangle with leg $2\sqrt{2}$ opposite $45^\circ$, hypotenuse $x$. 3. **Formula:** In a $45^\circ-45^\circ-90^\circ$ triangle, legs are equal and hypotenuse $= \text{leg} \times \sqrt{2}$. 4. **Apply formula:** $$x = 2\sqrt{2} \times \sqrt{2}$$ 5. **Simplify:** $$x = 2 \times \cancel{\sqrt{2}} \times \cancel{\sqrt{2}} = 2 \times 2 = 4$$ 6. **Answer for first triangle:** $x=4$. 7. **Second triangle:** Right triangle with angle $30^\circ$, adjacent side $16$, opposite side $x$. 8. **Formula:** In a $30^\circ-60^\circ-90^\circ$ triangle, opposite $30^\circ$ side $= \frac{1}{\sqrt{3}}$ times adjacent side to $30^\circ$ (or use tangent). 9. **Use tangent:** $$\tan 30^\circ = \frac{x}{16}$$ 10. **Value:** $$\tan 30^\circ = \frac{1}{\sqrt{3}}$$ 11. **Solve for $x$:** $$x = 16 \times \frac{1}{\sqrt{3}} = \frac{16}{\sqrt{3}}$$ 12. **Rationalize denominator:** $$x = \frac{16}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{16\sqrt{3}}{3}$$ 13. **Answer for second triangle:** $x=\frac{16\sqrt{3}}{3}$. 14. **Summary:** - First triangle: $x=4$ - Second triangle: $x=\frac{16\sqrt{3}}{3}$