http://screensnapr.com/v/zn5DFe.png No idea.

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http://screensnapr.com/v/zn5DFe.png No idea.

Mathematics
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use the cosine rule, u have the required lengths, the only thing missing is the angle, plug in the values into the equation and solve for \(\theta\)
what saso said. law 'o cosines for this one
a^2=b^2+c^2-2(b)(c)cosA

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So then..
yes but of course you want \[\theta\] not a length
9=2^2+3^2-2(2)(3)cosX
what you are looking for is the angle right? so you need to use \[\theta=\cos^{-1}(\frac{b^2+c^2-a^2}{2bc})\]
or you could use what you wrote above to solve for \[\cos(\theta)\] and then take the inverse cosine to get theta. amounts to the same thing
\(a=4\frac{1}{2} \ \ b=2 \ \ \ c=3\)
I got 70.5
One last question! Woohoo!
looking at the picture, u can simply tell that the angle is greater than 90... 70.5 is definitely not the right answer
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