Yes, the laws apply to right-angled triangles as well. But, they're not particularly interesting there:
For $\triangle ABC$ with $\theta = \angle ABC$ a right angle, we can try to apply the cosine law about the right angle, and get $AC^2 = AB^2 + BC^2 - AB\cdot BC\cdot \cos\theta = AB^2 + BC^2$, as $\cos 90^\circ$ = 0. But this is nothing more than Pythagoras' theorem!
For the sine law, letting $\alpha = \angle ACB$ and $\beta = \angle BAC$, we have that
$$\begin{align}
\frac{\sin \theta}{AC} &= \frac{\sin \alpha}{AB} = \frac{\sin \beta}{BC}\\
\frac{1}{AC} &= \frac{\sin \alpha}{AB} = \frac{\sin \beta}{BC} & \Leftrightarrow\\
{\sin \alpha} &= \frac{AB}{AC}\\
{\sin \beta} &= \frac{BC}{AC},
\end{align}$$
which is nothing more than the definition of the sine of an angle.
So you can think of the rules for right-angled triangles (Pythagoras' theorem and the trigonometric relationships) as special cases of more general formulae. These special cases are typically easier to use than the general formula, so that's why they're used wherever possible.