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#### xyz_1965

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- Jul 26, 2020

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- Jan 30, 2018

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The "law of tangents" (I had to look it up) says that

"if a is the side of a triangle opposite angle α and b is the side of a triangle opposite angle β then (a-b)/(a+b)=(tan(1/2(α-β)))/(tan(1/2(α+β)))".

You are right- that is not very often taught. I suspect that is because it is much more complicated, especially the dependence on half angles, than either the sine or cosine laws and that it can be replace by those laws.

"if a is the side of a triangle opposite angle α and b is the side of a triangle opposite angle β then (a-b)/(a+b)=(tan(1/2(α-β)))/(tan(1/2(α+β)))".

You are right- that is not very often taught. I suspect that is because it is much more complicated, especially the dependence on half angles, than either the sine or cosine laws and that it can be replace by those laws.

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Someone deleted 3 or 4 threads I posted about 2 days ago. The reason given was NO WORK SHOWN. I can only show work if I know how to start the solution of the problem.The "law of tangents" (I had to look it up) says that

"if a is the side of a triangle opposite angle α and b is the side of a triangle opposite angle β then (a-b)/(a+b)=(tan(1/2(α-β)))/(tan(1/2(α+β)))".

You are right- that is not very often taught. I suspect that is because it is much more complicated, especially the dependence on half angles, than either the sine or cosine laws and that it can be replace by those laws.

All I needed was a simple hint or step to get me started. The questions did not need to be deleted. Again, there are questions in my precalculus textbook for which the author dooes not provide samples.

I agree that the Law of Tangent is not needed if the other two solve the problem. Be that as it may, it looks interesting. Can you solve a problem using the Law of Tangent? I just want to see what it looks like. Can you prove the Law of Tangent?

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$\frac{a- b}{a+ b}= \frac{tan(\frac{1}{2}(\alpha- \beta))}{tan(\frac{1}{2}(\alpha+ \beta)}$.

xyz- 1965: I wondered why there was no specific question in your post.

Seriously? If you are given homework where you do not even know how toSomeone deleted 3 or 4 threads I posted about 2 days ago. The reason given was NO WORK SHOWN. I can only show work if I know how to start the solution of the problem.

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That would be me.Someone deleted 3 or 4 threads I posted about 2 days ago. The reason given was NO WORK SHOWN. I can only show work if I know how to start the solution of the problem.

All I needed was a simple hint or step to get me started. The questions did not need to be deleted. Again, there are questions in my precalculus textbook for which the author dooes not provide samples.

Please read the forum rules, which is what we enforce.

The relevant rule is:

Show some effort. If you want help with a question we expect you to show some effort. Effort might include showing your work, learning how to typeset equations using $\LaTeX$, making your question clearer, titling threads effectively and posting in the appropriate subforum, making a genuine attempt to understand the given help before asking for more help, and learning from previously asked questions. Moderators reserve the right to close threads in cases where the member is not making a genuine effort (particularly if the member is flooding the forums with multiple questions of the same type). You also should remember that all contributors to MHB are unpaid volunteers and are under no obligation to answer a question.

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I somehow missed that...I'm trying to encourage the OP to use LaTeX and so my post wasn't meant to address your post although I can see it looked that way.MarkFL: I believe I stated it above but here it is in Latex:

$\frac{a- b}{a+ b}= \frac{tan(\frac{1}{2}(\alpha- \beta))}{tan(\frac{1}{2}(\alpha+ \beta)}$.

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Yo Mark,Can you state this law, preferably using \(\displaystyle \LaTeX\) for readability?

How is LaTex done? I have no idea.

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1. I am 55 not 15 in high school.MarkFL: I believe I stated it above but here it is in Latex:

$\frac{a- b}{a+ b}= \frac{tan(\frac{1}{2}(\alpha- \beta))}{tan(\frac{1}{2}(\alpha+ \beta)}$.

xyz- 1965: I wondered why there was no specific question in your post.

Seriously? If you are given homework where you do not even know how tostartthen you need to tell your teacher that! Your teacher needs to know to give you additional teaching.

2. What? Don't believe me? Ask MarkFL. He knows more about me than everyone here.

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Why are you attacking me? I told you millions of times that I do not know how to use Latex.I somehow missed that...I'm trying to encourage the OP to use LaTeX and so my post wasn't meant to address your post although I can see it looked that way.

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How am I attacking you?Why are you attacking me? I told you millions of times that I do not know how to use Latex.

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I do not know Latex. You want me to post using LaTex.How am I attacking you?

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Ok. Can you check my recent posts to see if it is correct?

If the next chapter in your personal studies is the Law of Sines and Law of Cosines, how can you agree that the Tangent law is not needed.I agree that the Law of Tangent is not needed if the other two solve the problem. Be that as it may, it looks interesting. Can you solve a problem using the Law of Tangent? I just want to see what it looks like.

PrecalculusWithUnit-CircleTrigonometry4e2006CohenLeeSkylar,page726

TrigonometricDelights1998EliMaor,page152

Scaum'sOutlineOfTheory&ProblemsOfPlane&SphericalTrigonometry1954FrankAyres,Jr.,page105

Now that you've seen what it looks like using tables, would you actually care to have a go at it using your calculator? Ah yes, I forget. You're that passionate math hobbyist who keeps posting math problems but doesn't own a calculator. Or was that your other split personality harpazzo. It's getting hard to keep track of your split personalities.

At any rate, if you do finally get a calculator, you need but note that A + B + C = 180° and you're basically done. A bit of creative thinking and blind application of the Tangent law formula should satisfy your wandering curiosity.