Lukecrayonz
  • Lukecrayonz
http://screensnapr.com/v/zn5DFe.png No idea.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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sasogeek
  • sasogeek
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 \(\theta\)
anonymous
  • anonymous
what saso said. law 'o cosines for this one
Lukecrayonz
  • Lukecrayonz
a^2=b^2+c^2-2(b)(c)cosA

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