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I am supposed to find the value of $\theta$ in a triangle. I am given a triangle with 3 lengths and no angles. The bottom left is $\theta$, bottom line is 5, right is 7 and left is 3. I have no idea what to do. My book used some strange forumla $\cos(\theta)= (b^2 + c^2 - a^2)/2bc$.

How do I know what is a $b$ and $c$? i tried it and got the wrong answer, only one combination of abc works. I need to memorize this forumla for a test tomorrow, along with the 3 other cosine laws, and the 3 sin laws and the 21 other half and double identities, and the cofunctions.

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closed as too localized by t.b., yunone, Andres Caicedo, Arturo Magidin, Qiaochu Yuan Jul 3 '11 at 2:28

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I'm voting to close this question. Adam: you asked 35 questions and still are unable to typeset your questions. You still haven't registered and haven't upvoted a single answer you were given. You're using strong language offending many people here, repeatedly, and after being told many times to cut it out. I find this attitude unbearable. –  t.b. Jul 1 '11 at 0:46
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Thanks friends, I guess I will just fail this class and switch degrees. –  Adam Jul 1 '11 at 0:48
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Dear Adam, please change your attitude. If you are seriously thinking about failing this class and switching degrees simply if you do not receive help here, then this is worrying. –  Amitesh Datta Jul 1 '11 at 1:25
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Well if I fail the test tomorrow (which is very likely) I am going to fail the class and lose around $2000 for having to pay the class back, and then taking it over again. I am attempting to study for the test but I am literally stuck on this part and can't get any further. I know you have absolutely no idea what it is like to have to study incredibly hard at something and then constantly fail at it, but it sucks and it sucks even more when there is a lot of money depending on passing it. Especially when that money is what seperates me from being homeless and probably ending up dead. –  Adam Jul 1 '11 at 1:30
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"Thanks friends, I guess I will just fail this class tomorrow." I'm not wild about the passive-aggressiveness, nor is it the first time we see it shining through; in fact, I find it, in view of the amount of help that has been offered, more than just a bit insulting. –  Arturo Magidin Jul 1 '11 at 3:21

1 Answer 1

up vote 5 down vote accepted

The reason only one combination of $a$, $b$, and $c$ works is that the labeling of the triangle is important. The law of cosines says that, in the triangle

enter image description here

we have

$$\cos(\gamma)=\frac{a^2+b^2-c^2}{2ab}.$$

The side of length $a$ is the side that is opposite the angle $\alpha$, and similarly with side $b$ and angle $\beta$, and side $c$ and angle $\gamma$. (Note that $\alpha$, $\beta$, and $\gamma$ are the first three letters in the Greek alphabet, and they are paired respectively with $a$, $b$, and $c$.)

The formulas

$$\cos(\gamma)=\frac{b^2+c^2-a^2}{2bc}$$ and $$\cos(\gamma)=\frac{a^2+c^2-b^2}{2ac}$$ are false, because they are using the wrong combination of sides and angles. The formulas $$\cos(\alpha)=\frac{b^2+c^2-a^2}{2bc}$$ and $$\cos(\beta)=\frac{a^2+c^2-b^2}{2ac}$$ are the true versions.

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I still don't get it, on a triangle with only theta as an angle, how do I determine what A B and C are? I am guessing that theta and A should pair together but that gives an incorrect answer. –  Adam Jul 1 '11 at 0:58
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$\theta$ goes with whatever side it is opposite. From your description, it sounds like the side of length $7$ is opposite $\theta$. So $$\cos(\theta)=\frac{(\text{1st other side})^2+(\text{2nd other side})^2-(\text{side opposite }\theta)^2}{2(\text{1st other side})(\text{2nd other side})}=$$ $$\frac{5^2+3^2-7^2}{2(5)(3)}=-\frac{1}{2}$$ so that $\theta=\arccos(-\frac{1}{2})=\frac{2\pi}{3}$, or $120^\circ$. –  Zev Chonoles Jul 1 '11 at 1:13
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That is what I did and I got the wrong answer somehow. –  Adam Jul 1 '11 at 1:14
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@Adam: What did you get, and what do you think the correct answer is? –  Zev Chonoles Jul 1 '11 at 1:17
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Correct. It is unfortunate that this is the notation, but when dealing with trigonometric functions, the ${}^{-1}$ does not mean "one over", it means "inverse". So, for example, $$\sin^{-1}\qquad\text{ really means }\qquad\arcsin$$ $$\cos^{-1}\qquad\text{ really means }\qquad\arccos$$ $$\tan^{-1}\qquad\text{ really means }\qquad\arctan$$ –  Zev Chonoles Jul 1 '11 at 1:52

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