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Lukecrayonz

  • 4 years ago

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

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  1. sasogeek
    • 4 years ago
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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 \(\theta\)

  2. anonymous
    • 4 years ago
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    what saso said. law 'o cosines for this one

  3. Lukecrayonz
    • 4 years ago
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    a^2=b^2+c^2-2(b)(c)cosA

  4. Lukecrayonz
    • 4 years ago
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    So then..

  5. anonymous
    • 4 years ago
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    yes but of course you want \[\theta\] not a length

  6. Lukecrayonz
    • 4 years ago
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    9=2^2+3^2-2(2)(3)cosX

  7. anonymous
    • 4 years ago
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    what you are looking for is the angle right? so you need to use \[\theta=\cos^{-1}(\frac{b^2+c^2-a^2}{2bc})\]

  8. anonymous
    • 4 years ago
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    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

  9. sasogeek
    • 4 years ago
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    \(a=4\frac{1}{2} \ \ b=2 \ \ \ c=3\)

  10. Lukecrayonz
    • 4 years ago
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    I got 70.5

  11. Lukecrayonz
    • 4 years ago
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    One last question! Woohoo!

  12. sasogeek
    • 4 years ago
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    looking at the picture, u can simply tell that the angle is greater than 90... 70.5 is definitely not the right answer

  13. sasogeek
    • 4 years ago
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