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highschoolmom2010

  • 2 years ago

Find the values of w and then x. Round lengths to the nearest tenth and angle measures to the nearest degree.

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  1. highschoolmom2010
    • 2 years ago
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  2. katherine.ok
    • 2 years ago
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    you have multiple ways. one is to recognize that this is isoceles triangle ( the triangle shape with two equal sides)

  3. highschoolmom2010
    • 2 years ago
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    i saw that to begin with

  4. katherine.ok
    • 2 years ago
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    which makes the right triangle have internal sum of 90+ 42+w= 180.

  5. katherine.ok
    • 2 years ago
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    or does w mean the height?

  6. highschoolmom2010
    • 2 years ago
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    i believe w is height

  7. highschoolmom2010
    • 2 years ago
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    im not entirely sure

  8. katherine.ok
    • 2 years ago
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    then sin42= w/102

  9. katherine.ok
    • 2 years ago
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    and cos 42= (2/x)/102

  10. katherine.ok
    • 2 years ago
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    cos42= (x/2)/102****

  11. highschoolmom2010
    • 2 years ago
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    \[(102)*\sin 42=\frac{ w }{ 102 }*102\] \[102 \sin 42=w\] w\[\approx 68\]

  12. highschoolmom2010
    • 2 years ago
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    so how now that i have found w how do i get x?

  13. katherine.ok
    • 2 years ago
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    cos42= (x/2)/102

  14. anonymous
    • 2 years ago
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    pythagoras

  15. anonymous
    • 2 years ago
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    half of \(x\) is \(\sqrt{102^2-68^2}\)

  16. highschoolmom2010
    • 2 years ago
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    can i have a fraction in the numerator

  17. anonymous
    • 2 years ago
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    so \(x=2\sqrt{102^2-68^2}\)

  18. anonymous
    • 2 years ago
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    http://www.wolframalpha.com/input/?i=2sqrt%28102^2-68^2%29

  19. jdoe0001
    • 2 years ago
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    @highschoolmom2010 you can always just use the cosine of 42 degrees you know don't spare functions with your SOH CAH TOA

  20. katherine.ok
    • 2 years ago
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    Satelite using wolfram alpha. LOL hes definitely not highschool student.

  21. katherine.ok
    • 2 years ago
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    cos42=x/204= (x/2)/102 (the one i typed few mins ago)

  22. highschoolmom2010
    • 2 years ago
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    ^^confused -_-

  23. highschoolmom2010
    • 2 years ago
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    @ganeshie8 any help

  24. jdoe0001
    • 2 years ago
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    \(\bf sin(42^o) = \cfrac{\textit{opposite side}}{\textrm{hypotenuse}} \implies \cfrac{w}{102} \implies \color{blue}{sin(42^o) \times 102} = w\\ cos(42^o) = \cfrac{\textit{adjacent side}}{\textrm{hypotenuse}} \implies \cfrac{\textit{half of "x"}}{102} \implies \color{green}{cos(42^o) \times 102} = \textit{half of "x"}\)

  25. highschoolmom2010
    • 2 years ago
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    \[\cos (42)\rightarrow 65*102\]

  26. jdoe0001
    • 2 years ago
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    well, if "half of x" = 65, yes

  27. highschoolmom2010
    • 2 years ago
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    ive done confused myself again

  28. jdoe0001
    • 2 years ago
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    then again, if you already know that x= 65 * 2, you don't have to do the cosine part :)

  29. highschoolmom2010
    • 2 years ago
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    aw heck imma use @satellite73 s method

  30. highschoolmom2010
    • 2 years ago
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    \[\sqrt{108^2+68^2}=108-68=40\] \[\sqrt{40}=2\sqrt{10}\]

  31. highschoolmom2010
    • 2 years ago
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    but that doesnt look right

  32. jdoe0001
    • 2 years ago
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    using the cosine, I get ahemm, something else for "x"

  33. highschoolmom2010
    • 2 years ago
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    i dont understand now to use it with this problem

  34. highschoolmom2010
    • 2 years ago
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    for x

  35. katherine.ok
    • 2 years ago
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    okay wild katherine has appeared.

  36. highschoolmom2010
    • 2 years ago
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    |dw:1376262142492:dw|

  37. katherine.ok
    • 2 years ago
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    cos42=x/204

  38. highschoolmom2010
    • 2 years ago
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    :( math is really hard for me right now yall

  39. katherine.ok
    • 2 years ago
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    x=204*cos42.

  40. highschoolmom2010
    • 2 years ago
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    ya know that makes it a whole lot easier

  41. katherine.ok
    • 2 years ago
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    now the number should be above 150, which is so much more amazing than just sqrt(40).

  42. highschoolmom2010
    • 2 years ago
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    151.6

  43. highschoolmom2010
    • 2 years ago
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    ty @katherine.ok @satellite73 @jdoe0001

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