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Do the Law of Sines and the Law of Cosines apply to all triangles? Particularly, could you use these laws on right triangles?

That is, could you use these laws instead of the Sine=opposite/hypotenuse, Cosine=adjacent/hypotenuse, and Tangent=opposite/adjacent rules to solve right triangles?

I can't find this stated in any of my textbooks, nor has my instructor said anything about it, which I find odd.

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2 Answers

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The law of cosines applied to right triangles is the Pythagorean theorem, since the cosine of a right angle is $0$. $$ a^2 + b^2 - \underbrace{2ab\cos C}_{\begin{smallmatrix} \text{This is $0$} \\[3pt] \text{if } C\,=\,90^\circ. \end{smallmatrix}} = c^2. $$

Of course, you can also apply the law of cosines to either of the other two angles.

$$\text{sine}=\frac{\text{opposite}}{\text{hypotenuse}};\text{ therefore }\frac{\text{opposite}}{\text{sine}} =\frac{\text{hypotenuse}}{1} = \frac{\text{hypotenuse}}{\sin90^\circ}.$$ Therefore, the law of sines applied to right triangles is valid.

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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.

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