1. **Problem 17:** Given a right triangle with a 45° angle, hypotenuse = 4, legs = x and y. 2. Since the triangle is right-angled and has a 45° angle, it is an isosceles right triangle, so legs are equal: $x = y$. 3. Use the Pythagorean theorem: $$x^2 + y^2 = 4^2$$ 4. Substitute $y = x$: $$x^2 + x^2 = 16$$ 5. Simplify: $$2x^2 = 16$$ 6. Divide both sides by 2: $$\cancel{2}x^2 = \cancel{2}8$$ 7. So, $$x^2 = 8$$ 8. Take the square root: $$x = \sqrt{8} = 2\sqrt{2}$$ 9. Since $y = x$, $$y = 2\sqrt{2}$$ --- 1. **Problem 18:** Right triangle with 45° angle, hypotenuse = x, one leg = $3\sqrt{2}$, other leg = y. 2. For 45° right triangle, legs are equal, so $$y = 3\sqrt{2}$$ 3. Use Pythagorean theorem: $$ (3\sqrt{2})^2 + y^2 = x^2$$ 4. Substitute $y = 3\sqrt{2}$: $$ (3\sqrt{2})^2 + (3\sqrt{2})^2 = x^2$$ 5. Calculate squares: $$9 \times 2 + 9 \times 2 = x^2$$ 6. Simplify: $$18 + 18 = x^2$$ 7. So, $$x^2 = 36$$ 8. Take square root: $$x = 6$$ --- 1. **Problem 19:** Right triangle with 30° angle, hypotenuse = x, one leg = 2, other leg = y. 2. In a 30°-60°-90° triangle, the side opposite 30° is half the hypotenuse: $$2 = \frac{x}{2}$$ 3. Multiply both sides by 2: $$\cancel{2} \times 2 = \cancel{2} \times \frac{x}{2}$$ 4. So, $$4 = x$$ 5. The side opposite 60° (y) is $$y = x \times \frac{\sqrt{3}}{2}$$ 6. Substitute $x=4$: $$y = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3}$$ --- **Final answers:** - Problem 17: $x = 2\sqrt{2}$, $y = 2\sqrt{2}$ - Problem 18: $x = 6$, $y = 3\sqrt{2}$ - Problem 19: $x = 4$, $y = 2\sqrt{3}$