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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.

Michael Hardy
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    if you write them out carefully, the two Laws you mention, used on a right triangle, give you back the ratios you already know, " Sine=opposite/hypotenuse, Cosine=adjacent/hypotenuse, and Tangent=opposite/adjacent." I suggest you write these out. I think Cosines for one of the non-right angles may give something that appears a little different. – Will Jagy May 19 '14 at 18:04
  • In particular, the law of cosines for right triangles reduces to just the familiar Pythagorean theorem. – David H May 19 '14 at 18:06
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    Indeed they do, but because sine and cosine are functions defined (in a sense) using right triangles, they become largely redundant in the cases of right triangles. – Pockets May 19 '14 at 18:07
  • did it, Cosines for the right angle does give Pythagoras; if you then do Cosines for a different angle and substitute in Pyth., you get back cosine = adjacent/hypot. – Will Jagy May 19 '14 at 18:10
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    Since the OP said "all triangles", perhaps it's worth mentioning that the situation is a little different for spherical and hyperbolic ones. – sourisse May 19 '14 at 18:31
  • @Sourisse : yes you are absolutely right, so I'll mention it :) That it holds Only for triangles in Euclidean geometry. They are not always true for triangles on the surface of sphere for example. – jimjim Jul 28 '15 at 06:44

2 Answers2

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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.

Michael Hardy
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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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