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Math Help - Angles

  1. #1
    ADY
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    Angles

    Ok - So 2 ends of a cord of length 5.6m are attached to a beam 4.8m apart. A shoe is hanging off the cord 2.1m from one end.

    When i'm calculating the angles that are the 2 sections of cord meet the beam, is this correct...

    Using the cosine rule, we can asign a = 5.6, b = 4.8, c = 2.1

    So...

    CosA = \frac{4.8^2 + 2.1^2 - 5.6^2}{2 * 4.8 * 2.1} = -0.193948

    So arccos(-0.193948) = 101.18

    I'd then do the same to find the other angle?

    If not please could someone advise

    Last edited by ADY; December 6th 2009 at 11:56 PM. Reason: LaTEx
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by ADY View Post
    Ok - So we have a cord in length 5.6m, which are attached to a length 4.8m apart. A shoe is hanging off the rope 2.1m from one end.

    When i'm calculating the angles that are the 2 sections of cord meet the length, is this correct...

    Using the cosine rule, we can asign a = 5.6, b = 4.8, c = 2.1

    So...

    CosA = \frac{4.8^2 + 2.1^2 - 5.6^2}{2 * 4.8 * 2.1} = -0.193948

    So arccos(-0.193948) = 101.18


    I'd then do the same to find the other angle?

    If not please could someone advise

    correct then you can use sine low for less calculations

    \frac{a}{\sin A} = \frac{b}{\sin B}
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  3. #3
    ADY
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    But surely i'll end up with 3 angles and i only need two?

    so where do i go from here?
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by ADY View Post
    But surely i'll end up with 3 angles and i only need two?

    so where do i go from here?
    I solved it, since I think we have a triangle

    the question said a chord 5.6m attached to a another one with length 4.8 right ??
    then a shoe is hanging off the rope 2.1 m from one end

    the end of what ?? and the shoe is hanging off which rope you said before chord then rope ??

    can you explain
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  5. #5
    ADY
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    I make 3 angles:

    101.8 , 57.23, 21.58

    but i dont understand which two angles relate to the rope coming away from the beam
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  6. #6
    ADY
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    after the outcome to the above i need to draw a force diagram and then a triangle of forces where the shoe weighs 36g..any ideas?
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  7. #7
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by ADY View Post
    Ok - So 2 ends of a cord of length 5.6m are attached to a beam 4.8m apart. A shoe is hanging off the cord 2.1m from one end.

    finally I get it this is the graph, okay all things are clear what the question want can you tell me

    Angles-finally.jpg
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  8. #8
    ADY
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    Thats the one

    So trying to figure out the angles that meet the 4.8 line...
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  9. #9
    MHF Contributor Amer's Avatar
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    call the angle between 2.1 and 3.5 A so the side 4.8=a the side 2.1=b and the side 3.5=c

    as you solved before by cosine low

    b^2 = a^2 + c^2 - 2ac\cdot \cos B

    (2.1)^2 = (4.8)^2 + (3.5)^2 - 2(3.5)(4.8) \cos B

    \cos B = \frac{4.8^2 + 3.5^2 - 2.1^2 }{2(3.5)(4.8)}

    B = \cos ^{-1} \left(\frac{4.8^2 + 3.5^2 - 2.1^2 }{2(3.5)(4.8)}\right)

    you can continue
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  10. #10
    ADY
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    So we're saying b = 23.07

    So do we just repeat that same process for c ?
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  11. #11
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by ADY View Post
    So we're saying b = 23.07

    So do we just repeat that same process for c ?

    ya or you can use sine law
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  12. #12
    ADY
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    so C = 40.78 ?
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  13. #13
    ADY
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    and therefore A = 117.15?

    please could you confirm my results
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  14. #14
    ADY
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    How can this relate to a forces diagram?
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  15. #15
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by ADY View Post
    How can this relate to a forces diagram?
    can you write the whole question ?
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