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Please help me!

  1. Jun 1, 2009 #1
    1. The problem statement, all variables and given/known data

    Staff and students at a school have been complaining about the location of the water fountains. You must decide if one or both fountains need to be relocated.

    Calculate the measure of angle 0, formed by the "line of sight" from the middle of the double doorway to the water fountains on the opposite wall.

    2. Relevant equations

    http://img261.imageshack.us/img261/1636/fountain1.png [Broken]

    3. The attempt at a solution

    am I supposed to use the sine law or the cosine law or something like that? or sohcahtoa ratio?

    plz explain to me how I solve these step-by-step so I can understand it and learn from it for my quiz tomorrow, thanks!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 1, 2009 #2

    rock.freak667

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    So you have a triangle, with three sides of which you know the lengths.
    You can only use the ratios of sine,cosine and tangent when you have a right angle. So can you use soh,cah,toa ratios?

    If you can, then use it. If not consider the alternative sine and cosine laws.
    Write out the formulas for these and see which one is most useful for this question by checking the variables you know.
     
  4. Jun 1, 2009 #3
    it must be the cosine law, right? how do i go about finding the angle using the cosine law?
     
  5. Jun 1, 2009 #4

    rl.bhat

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    If opposite side of the angle θ is R and other two sides are P and Q, then
    R^2 = P^2 + Q^2 - 2PQ*cosθ.
     
  6. Jun 1, 2009 #5

    Ouabache

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    Are you supposed to infer from your description where the present location of the water fountains are?
    What is the cosine law?
     
  7. Jun 1, 2009 #6
    i got the cosine law for finding angles its:

    cosA = b2 + c2 - a2

    and divide all of that by 2bc
     
  8. Jun 1, 2009 #7

    symbolipoint

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    The answer is represented in this response:
    Just identify the corresponding parts.
     
  9. Jun 1, 2009 #8
    but im not finding the length, arent i? I'm supposed to find the angle. I thought that equation above is for finding lengths.

    opposite side of angle 0 would be c, and other two sides would be a and b
     
  10. Jun 1, 2009 #9

    symbolipoint

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    Again, what are the corresponding parts? Substitute the corresponding parts into the formula. What quantities are known, and what quantities are unknown? The rest is simple algebra and a small amount of basic Trigonometry.
     
  11. Jun 1, 2009 #10
    R^2 = P^2 + Q^2 - 2PQ*cosθ.

    465^2 = P^302 + Q^237 - 2(302)(237)*cosθ

    is it like that?

    can I also use this formula:



     
  12. Jun 2, 2009 #11

    rl.bhat

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    465^2 = P^302 + Q^237 - 2(302)(237)*cosθ
    It should be
    465^2 = 302^2 + 237^2 - 2(302)(237)*cosθ.
    You can use the other formula also.
     
  13. Jun 2, 2009 #12

    HallsofIvy

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    It would be better to use "^" to indicate powers and put things in parentheses:
    cos A= (b^2+ c^2- a^2)/(2bc), but yes, that is what every one has been trying to tell you: start from the standard form of the cosine law and solve for cos A. Or put the numbers given into a^2= b^2+ c^2- 2bc cos(A) first and solve for cos A.

    As rl.bhat told you, it would be 465^2= 302^2+ 237^2- 2(302)(237)cos(A), not your
    "465^2= P^302+ Q^237- 2(302)(237)cos(A)". I assume that was just "temporary insanity"!
     
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