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I've discovered a new method to solve SAS triangles without using the law of cosines:

In $\triangle ABC$, if sides $AC$ and $BC$, and angle $C$, are known, then: $$\tan B =\frac{AC \sin C}{BC- AC \cos C}$$

Putting in mind that I'm still a preparatory school student, which means that this is a great achievement for me, should I write a paper about this? or would it be a waste of time?

Blue
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Ahmed M. Elsonbaty
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  • @jkabrg I know but I found another formula which can be used instead of the law of cosines (and the proof doesn't include the use of the law of cosines or sines) – Ahmed M. Elsonbaty Nov 26 '19 at 17:20
  • Can you give a proof (or sketch of one) for your formula? Also I assume the x in the top is just "multiply", is that right? – coffeemath Nov 26 '19 at 17:21
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    Congratulations on the discovery, both of the result itself and of the satisfaction mathematicians feel each time this happens. ... The result nicely abbreviates a bit of work that, eg, @MatthewDaly's answer shows, but since that work is pretty straightforward, the general benefit is reduced: it might take more effort to memorize your additional result than to just duplicate the steps. (In practice, a proof in a journal article or somesuch would typically skip these kinds of details anyway, leaving the reader to fill in gaps.) While I wouldn't consider this paper-worthy, it's pretty neat. – Blue Nov 26 '19 at 17:51
  • A minor suggestion: such expressions almost always looks better with angles $\alpha,\beta,\gamma$ and sides $a,b,c$. – g.kov Nov 26 '19 at 19:45

2 Answers2

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Not groundbreaking, but well observed!

enter image description here

With $\overline{AD}\perp\overline{BC}$,

$$\tan B=\frac{AD}{BD}=\frac{AD}{BC-CD}=\frac{AC\sin C}{BC-AC\cos C}$$

Community
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  • and that's the same method I used."Not groundbreaking, but well observed!" ok I won't publish it but I will just leave it here – Ahmed M. Elsonbaty Nov 26 '19 at 17:37
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    @AhmedM.Elsonbaty The discouraging thing is that all of the easy math has already been discovered. ^_^ But this is the same process of discovery that will persist throughout a mathematical career! –  Nov 26 '19 at 17:39
  • Beat me by a minute; +1 anyways. – Michael Hoppe Nov 26 '19 at 17:42
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Take the law of sines $$\frac{a}{\sin A}=\frac{b}{\sin B}$$ and rearrange to get $$\frac{a}{b}=\frac{\sin A}{\sin B}.$$

Divide both sides by $\sin C$ to get $$\frac{a}{b\sin C}=\frac{\sin A}{\sin B\sin C}$$ and write $A$ in terms of $B$ and $C$ to get $$\frac{a}{b\sin C}=\frac{\sin(\pi-(B+C))}{\sin B\sin C}=\frac{\sin (B+C)}{\sin B\sin C}.$$

Using an identity, we get $$\frac{\sin(B+C)}{\sin B\sin C}=\csc B\csc C\sin (B+C)=\cot B+\cot C$$ and we have $$\cot B+\cot C=\frac{a}{b\sin C}.$$

Rearrange to get $$\cot B=\frac{a}{b\sin C}-\cot C$$ and write the RHS as a single rational expression: $$\cot B=\frac{a}{b\sin C}-\frac{\cos C}{\sin C}=\frac{a-b\cos C}{b\sin C}.$$

Finally, reciprocate both sides to get $$\tan B=\frac{b\sin C}{a-b\cos C}$$ as desired. This comes from the law of sines, after all.

Andrew Chin
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