Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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] ]

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

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

share|improve this question
1  
$x$ 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
1  
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

2 Answers 2

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 $x$.

share|improve this answer
    
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 $x=\sqrt{28} \approx 5.291$.

share|improve this answer
    
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
2  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.