1. The problem involves finding missing side lengths in various right triangles with given angles and side lengths. 2. We use the Pythagorean theorem and special right triangle ratios: for 45°-45°-90° triangles, sides are in ratio $1:1:\sqrt{2}$; for 30°-60°-90° triangles, sides are in ratio $1:\sqrt{3}:2$. 3. For the first triangle (top-left), given hypotenuse $13$ and angle $45^\circ$, legs $x$ and $y$ are equal, so $x = y = \frac{13}{\sqrt{2}} = 13\frac{\sqrt{2}}{2} = 13\sqrt{2}/2$. 4. For the second triangle (top-center), hypotenuse is $30\sqrt{2}$, angle $45^\circ$, legs $x$ and $y$ equal, so $x = y = \frac{30\sqrt{2}}{\sqrt{2}} = 30$. 5. For the third triangle (top-right), angle $60^\circ$, leg $3$, hypotenuse $y$, so $y = \frac{3}{\sin 60^\circ} = \frac{3}{\sqrt{3}/2} = 2\times 3/\sqrt{3} = 2\sqrt{3}$. 6. For the fourth triangle (middle-left), angle $30^\circ$, legs $x$ and $y$, hypotenuse unknown. Using ratios, if $x$ is opposite $30^\circ$, then $x = \frac{1}{2} \times \text{hypotenuse}$, $y = \frac{\sqrt{3}}{2} \times \text{hypotenuse}$. Given $x=17\sqrt{3}$ and $y=17$, this contradicts the ratio, so likely $x=17\sqrt{3}$ is adjacent side, $y=17$ opposite side, hypotenuse $= 2y = 34$. 7. For the fifth triangle (middle-center), isosceles right triangle with legs $10$, hypotenuse $10\sqrt{2}$, consistent with $x=y=10$. 8. For the sixth triangle (middle-right), angle $60^\circ$, legs $x$ and $y$, hypotenuse $50$, and $25\sqrt{3}$. Using ratios, if hypotenuse is $50$, then shorter leg $= 25$, longer leg $= 25\sqrt{3}$. 9. For the seventh triangle (bottom-left), angle $45^\circ$, leg $y=14$, so $x=y=14$, hypotenuse $= 14\sqrt{2}$. 10. For the eighth triangle (bottom-center-left), angle $30^\circ$, hypotenuse $24$, legs $x$ and $y$ satisfy $x=12$, $y=12\sqrt{3}$. 11. For the ninth triangle (bottom-center-right), angle $60^\circ$, hypotenuse $22\sqrt{3}$, legs $x=11$, $y=11\sqrt{3}$. 12. For the tenth triangle (bottom-right), angle $30^\circ$, hypotenuse $\sqrt{6}$, legs $x=\frac{\sqrt{6}}{2}$, $y=\frac{\sqrt{6}\sqrt{3}}{2} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}}$. 13. For the eleventh triangle, angle $45^\circ$, legs $x$ and $y$, hypotenuse $a\sqrt{2} = 10$, with $a=\sqrt{5}$, so legs $x=y=\sqrt{5}$. 14. For the twelfth triangle, angle $60^\circ$, hypotenuse $4\sqrt{21}$, legs $x=2\sqrt{7}$, $y=4\sqrt{21} \times \frac{\sqrt{3}}{2} = 2\sqrt{63} = 6\sqrt{7}$. 15. For the thirteenth triangle composed of two right triangles with angles $45^\circ$ and $30^\circ$, sides $x,y,z$ and $17$ given, solving yields $x=17$, $y=17$, $z=17\sqrt{2}$. 16. For the fourteenth triangle, angle $60^\circ$, legs $x,y$, base $27$, and $z$ labeled, using ratios and Pythagorean theorem, $x=13.5$, $y=13.5\sqrt{3}$, $z=27$. Final answers for missing values: - $x=13\sqrt{2}/2$, $y=13\sqrt{2}/2$ - $x=30$, $y=30$ - $x=3$, $y=6$ - $x=17\sqrt{3}$, $y=17$ - $x=10$, $y=10$ - $x=25$, $y=25\sqrt{3}$ - $x=14$, $y=14$ - $x=12$, $y=12\sqrt{3}$ - $x=11$, $y=11\sqrt{3}$ - $x=\frac{\sqrt{6}}{2}$, $y=\frac{3}{\sqrt{2}}$ - $x=\sqrt{5}$, $y=\sqrt{5}$ - $x=2\sqrt{7}$, $y=6\sqrt{7}$ - $x=17$, $y=17$, $z=17\sqrt{2}$ - $x=13.5$, $y=13.5\sqrt{3}$, $z=27$