1. **Problem statement:** We have a right triangle with angles 45°, 60°, and 90°. The side adjacent to the 45° angle is 2, the side opposite to 45° is $x$, the side opposite to 60° is $y$, and the hypotenuse is $z$. We need to find $x$, $y$, and $z$. 2. **Recall trigonometric ratios:** - For an angle $\theta$, $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$ - $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$ - $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$ 3. **Find $x$ using $\tan 45^\circ$:** $$\tan 45^\circ = 1 = \frac{x}{2}$$ Multiply both sides by 2: $$2 \times 1 = \cancel{2} \times \frac{x}{\cancel{2}} \Rightarrow 2 = x$$ 4. **Find $z$ using $\cos 45^\circ$:** $$\cos 45^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{2}{z} = \frac{\sqrt{2}}{2}$$ Multiply both sides by $z$: $$z \times \frac{\sqrt{2}}{2} = 2$$ Multiply both sides by 2: $$z \times \sqrt{2} = 4$$ Divide both sides by $\sqrt{2}$: $$\cancel{z} \times \frac{\sqrt{2}}{\cancel{\sqrt{2}}} = \frac{4}{\sqrt{2}} \Rightarrow z = \frac{4}{\sqrt{2}}$$ Rationalize denominator: $$z = \frac{4}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$$ 5. **Find $y$ using $\sin 60^\circ$:** $$\sin 60^\circ = \frac{y}{z} = \frac{\sqrt{3}}{2}$$ Multiply both sides by $z$: $$y = z \times \frac{\sqrt{3}}{2} = 2\sqrt{2} \times \frac{\sqrt{3}}{2}$$ Simplify: $$y = \cancel{2} \sqrt{2} \times \frac{\sqrt{3}}{\cancel{2}} = \sqrt{2} \times \sqrt{3} = \sqrt{6}$$ **Final answers:** $$x = 2$$ $$y = \sqrt{6}$$ $$z = 2\sqrt{2}$$