Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

highschoolmom2010

  • one year ago

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

  • This Question is Closed
  1. highschoolmom2010
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  2. katherine.ok
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    you have multiple ways. one is to recognize that this is isoceles triangle ( the triangle shape with two equal sides)

  3. highschoolmom2010
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i saw that to begin with

  4. katherine.ok
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    which makes the right triangle have internal sum of 90+ 42+w= 180.

  5. katherine.ok
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    or does w mean the height?

  6. highschoolmom2010
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i believe w is height

  7. highschoolmom2010
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    im not entirely sure

  8. katherine.ok
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    then sin42= w/102

  9. katherine.ok
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    and cos 42= (2/x)/102

  10. katherine.ok
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    cos42= (x/2)/102****

  11. highschoolmom2010
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[(102)*\sin 42=\frac{ w }{ 102 }*102\] \[102 \sin 42=w\] w\[\approx 68\]

  12. highschoolmom2010
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so how now that i have found w how do i get x?

  13. katherine.ok
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    cos42= (x/2)/102

  14. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    pythagoras

  15. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    half of \(x\) is \(\sqrt{102^2-68^2}\)

  16. highschoolmom2010
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can i have a fraction in the numerator

  17. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so \(x=2\sqrt{102^2-68^2}\)

  18. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    http://www.wolframalpha.com/input/?i=2sqrt%28102^2-68^2%29

  19. jdoe0001
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Satelite using wolfram alpha. LOL hes definitely not highschool student.

  21. katherine.ok
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    cos42=x/204= (x/2)/102 (the one i typed few mins ago)

  22. highschoolmom2010
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ^^confused -_-

  23. highschoolmom2010
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @ganeshie8 any help

  24. jdoe0001
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  25. highschoolmom2010
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\cos (42)\rightarrow 65*102\]

  26. jdoe0001
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well, if "half of x" = 65, yes

  27. highschoolmom2010
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ive done confused myself again

  28. jdoe0001
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  29. highschoolmom2010
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    aw heck imma use @satellite73 s method

  30. highschoolmom2010
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\sqrt{108^2+68^2}=108-68=40\] \[\sqrt{40}=2\sqrt{10}\]

  31. highschoolmom2010
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but that doesnt look right

  32. jdoe0001
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    using the cosine, I get ahemm, something else for "x"

  33. highschoolmom2010
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i dont understand now to use it with this problem

  34. highschoolmom2010
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    for x

  35. katherine.ok
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    okay wild katherine has appeared.

  36. highschoolmom2010
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1376262142492:dw|

  37. katherine.ok
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    cos42=x/204

  38. highschoolmom2010
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    :( math is really hard for me right now yall

  39. katherine.ok
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    x=204*cos42.

  40. highschoolmom2010
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ya know that makes it a whole lot easier

  41. katherine.ok
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    now the number should be above 150, which is so much more amazing than just sqrt(40).

  42. highschoolmom2010
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    151.6

  43. highschoolmom2010
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ty @katherine.ok @satellite73 @jdoe0001

  44. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.