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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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    Now what are the maximum and minimum values of $A$, $\cos A$ and $x$?2012-04-18

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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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    i made the cha$n$ge2012-04-17
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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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    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.2012-04-17