Solve for missing sides in right-angled triangles given hypotenuse and an angle, or adjacent side and an angle.

$$\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$$ $$\sin(45^\circ) = \frac{x}{5}$$ $$x = 5 \times \sin(45^\circ)$$ $$x = 5 \times \frac{\sqrt{2}}{2}$$ $$x = \frac{5\sqrt{2}}{2}$$ $$x \approx 3.536$$

$$\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$$ $$\cos(45^\circ) = \frac{y}{5}$$ $$y = 5 \times \cos(45^\circ)$$ $$y = 5 \times \frac{\sqrt{2}}{2}$$ $$y = \frac{5\sqrt{2}}{2}$$ $$y \approx 3.536$$

The values are: $\boxed{x = \frac{5\sqrt{2}}{2}, y = \frac{5\sqrt{2}}{2}}$

:
We have a right-angled triangle with the adjacent side to $60^\circ$ as 2. We need to find the hypotenuse $x$ and the opposite side $y$.

$$\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$$ $$\cos(60^\circ) = \frac{2}{x}$$ $$x = \frac{2}{\cos(60^\circ)}$$ $$x = \frac{2}{0.5}$$ $$x = 4$$

$$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$$ $$\tan(60^\circ) = \frac{y}{2}$$ $$y = 2 \times \tan(60^\circ)$$ $$y = 2 \times \sqrt{3}$$ $$y \approx 3.464$$

The values are: $\boxed{x = 4, y = 2\sqrt{3}}$

:
We have a right-angled triangle with a hypotenuse of 10 and an angle of $60^\circ$. We need to find the adjacent side $x$ and the opposite side $y$.

$$\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$$ $$\cos(60^\circ) = \frac{x}{10}$$ $$x = 10 \times \cos(60^\circ)$$ $$x = 10 \times 0.5$$ $$x = 5$$

$$\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$$ $$\sin(60^\circ) = \frac{y}{10}$$ $$y = 10 \times \sin(60^\circ)$$ $$y = 10 \times \frac{\sqrt{3}}{2}$$ $$y = 5\sqrt{3}$$ $$y \approx 8.660$$

The values are: $\boxed{x = 5, y = 5\sqrt{3}}$