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I wish I could provide an image but I'll explain the best way I can.

There is a triangle that is not a $90^\circ$ triangle. It has two sides measured at 8 and 6 (units not specified). The other side is $x$. There are no angles whose measures are given. How do I find $x$? We are doing a topic on law of sines.

Law of cosines

$\displaystyle \large a^2 = b^2 + c^2 - 2bc \space \cos A$

$\displaystyle \large x^2 = 6^2 + 8^2 - 2(6)(8) \space \cos A$

x = Sqrt[4 Cos[A] ]

x = 2 Sqrt[ Cos[A] ]

$\displaystyle \large x^2 = -96\space \cos A + 100$

$\displaystyle \large x = \sqrt{100-96\space \cos A}$

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$x$ should be $ x = \sqrt{100-96\cos A}$ and also As you see that x depends on angle $A$. It will give a range if you change angle $A$ . x will be in range $2<x<14$. – Mathlover Apr 17 '12 at 22:19
Where does that value come from?? – David Apr 17 '12 at 22:21
Please be careful that $100-96cosA$ is not equal to $4cosA$ . This is mathematically wrong what you did in your question. – Mathlover Apr 17 '12 at 22:25
Oh yeah that completely slipped my mind!! Thanks for the save lol – David Apr 17 '12 at 22:26
Now what are the maximum and minimum values of $A$, $\cos A$ and $x$? – Henry Apr 18 '12 at 1:50
up vote 0 down vote accepted

There is not enough information to solve this problem. To uniquely define $x$, you would need to know the angle opposite the side of length $x$ as any 2 triangles with a side of length 6, one of length 8, and a congruent angle between them would be congruent by SAS. Although, it would probably be more appropriate to use the law of cosines in this case to solve for $x$.

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Use the law of cosines? I'll show you what I tried in the edit. – David Apr 17 '12 at 22:07
I made the edit! :) Is it right? – David Apr 17 '12 at 22:16
@David Close. You dropped a minus sign. Also, you may want to call the angle measure $X$ to differentiate it from the length of the side. – Mike Apr 17 '12 at 22:51
Got it, is the edit right? – David Apr 17 '12 at 22:57
@David This length is a real value. $i$ shouldn't appear in it. It should be $\sqrt{100-96\cos X}$ – Mike Apr 17 '12 at 23:03

By the triangle inequality $$2 \lt x \lt 14.$$

Since there is not a right angle, you can exclude $x=10$ and $x=\sqrt{28} \approx 5.291$.

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Excuse my french, but.......huh? – David Apr 17 '12 at 21:38
The angle between the sides of length $8$ and $6$ could be almost nothing, in which case $x$ is almost as small as $8-6$, or anything up to $180^\circ$ in which case $x$ is almost $8+6$. But since there is not right angle you need to exclude $\sqrt{8^2 \pm 6^2}$. – Henry Apr 17 '12 at 21:41
Henry mentions that if you dont know any angles and if you only know 2 sides .$|a-b|<x<|a+b|$ where $a$ and $b$ given sides , $x$ is unknown side in the triaangle. You will need to define an angle to define x, otherwise it will be just a range as given in the answer. – Mathlover Apr 17 '12 at 21:46
@Mathlover So there is no exact value but just an estimation? – David Apr 17 '12 at 22:02
Yes, you have just a range with given 2 sides. Please imagine you hold 2 pencils that one is 6 cm other 8cm. We can create max 14 cm line or 2 cm will remain if we put on each other parallel . if you define an angle between pencils then you will have other side fixed. – Mathlover Apr 17 '12 at 22:09

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