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Cosine Problem

  1. Oct 3, 2005 #1
    Hey all -

    I'm going insane because I can't think of how to do this...

    I can't for the life of me get this straight... I have a triangle that I'm trying to solve for the hypotenuse, angle theta is 70 degrees and I know the adjacent leg is 7500. Cosine SHOULD solve for the hypotenuse, right (adjacent/hypotenuse)? I can't get it to solve for the hypotenuse.

    I set it up like this:

    cos70 = 7500/y

    I need to get the 7500 on the side with the cos70, but wouldn't I have to multiply the right by 1/7500 to cancel it on that side for y (hypotenuse)?

    Any help would be appreciated, I'm running into this over and over with my probelms.


    Thanks!
    -
    Morgan
     
  2. jcsd
  3. Oct 3, 2005 #2
    Just invert both sides:
    1/cos 70 = y/7500
    then solve for y
     
  4. Oct 3, 2005 #3
    I have to be missing something... because there is a similar problem where they basically (diff. numbers) make it (7500)cos70 = y
     
  5. Oct 3, 2005 #4
    I'm no expert but I did it like this..
    cos70 = 7500/x x= hypot

    cos70 = .3420

    .3420 = 7500 /x
    .3420x = 7500
    x = 7500 / .3420
    x = 21929.82456
     
  6. Oct 3, 2005 #5
    OK
    Here's another approach:
    You have it set up correctly.
    Multiply both sides by y to get the y out of the denominator on the RHS:
    y*cos 70 = 7500
    then solve for y.
    The "similar problem" is not the same as this one. This is just a couple of fractions that are making your head hurt. Think about how you would solve this:
    2 = 1/x
    you can either take the inverse of both sides to get
    1/2 = x
    or you can multiply both sides by x to get
    2x = 1
    and then solve for x.
    It's the exact same result.
    hope this helps.
     
  7. Oct 3, 2005 #6
    hmmm okay I'm seeing it is not the same as the other problem that I thought was the same... so your answers make sense.

    thanks guys!
     
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