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Math Help - Triangle line lengths problem

  1. #1
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    Triangle line lengths problem

    Please can anyone advise on the steps required to determine the lengths of the lines

    qs
    pr
    tr
    ts

    in the attached diagram

    Thanks

    Luke
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  2. #2
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    Re: Triangle line lengths problem

    Can't see/reach your diagram...can you post it properly?
    Or at least describe it?
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  3. #3
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    Re: Triangle line lengths problem

    Please find pdf attached

    Regards

    Luke
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  4. #4
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    Re: Triangle line lengths problem

    Thanks; couple questions:
    ts is part of qs, right? (you show ts)
    is P3 really pr? Not pt?

    Can you redo diagram:
    change P1, P2, P3 to a, b, c
    show points using capitals, lengths using small; like triangle ABC with sides a,b,c:
    Code:
    B
    
    a          c
    
    C         b             A
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  5. #5
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    Re: Triangle line lengths problem

    Well, had another look at your problem: only qs can be calculated.
    r can be anywhere on the line that you show it on;
    so, to calculate pr (and tr,ts), angle rps is required as a given.

    IF you re-submit a diagram with this added given, please make
    sure that ALL points (lines crossing) are labelled with capital letters;
    if not, I'll let ProveIt do the work!!
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  6. #6
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    Re: Triangle line lengths problem

    Hi Wilmer

    I have attached the diagram with amended labeling.

    I have highlighted in yellow the lengths I wish to calculate. I would appreciate any guidance. I just wish to know the steps of the calculations for each.

    Thanks

    Luke
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  7. #7
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    Re: Triangle line lengths problem

    for those that prefer not to download ...
    Attached Thumbnails Attached Thumbnails Triangle line lengths problem-triangle2.jpg  
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  8. #8
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    Re: Triangle line lengths problem

    Luke, I don't know what YOU know, like Law of Sines/Cosines and the likes.
    Plus what is this problem: homework from Geometry class?
    Anyway, I'll show how to calculate AE.

    Since angleFDE = 90 and angleFDA = 25, then angleADE = 90 - 25 = 65.
    So in triangleADE, we have AD = 50, DE = 60, angleADE = 65.
    Using Cosine Law:
    AE = SQRT[50^2 + 60^2 - 2(50)(60)COS(65)] = ~59.7

    To see what YOU can do, take triangleADF:
    AD = 50, angleFDA = 25 and angleFAD = 5
    Can you calculate lengths of DF and AF?

    Also, are you able to calculate angleDAB?
    There is enough info given (the 2 angles) to permit that calculation.
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  9. #9
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    Re: Triangle line lengths problem

    Thanks Wilmer

    I can see the calculation for DF and AF by drawing a line at right angles to AD to the angle DFA and then creating a calculation on the TAN of the angles (FDA and FAD) to determine the length of the perpendicular line to line AD. Knowing this I can the determine DF and FA.

    Angle DAB is 85 degrees (90 - 5).

    I can see how to get the answer for length AB but not the other unknowns.


    Regards

    Luke
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  10. #10
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    Re: Triangle line lengths problem

    Quote Originally Posted by LukeP20110907 View Post
    > I can see the calculation for DF and AF by drawing a line at right angles to AD to the angle DFA
    > and then creating a calculation on the TAN of the angles (FDA and FAD) to determine the length
    > of the perpendicular line to line AD. Knowing this I can the determine DF and FA.

    Well, easier to use Law of Sines right away:
    DF / SIN(5) = 50 / SIN(150) ; DF = 50SIN(5) / SIN(150) = ~8.716
    AF / SIN(25) = 50 / SIN(150) ; AF = 50SIN(25) / SIN(150) = ~ 42.262

    > Angle DAB is 85 degrees (90 - 5).

    Correct.

    > I can see how to get the answer for length AB but not the other unknowns.

    OK, show me how YOU'd do that...then we'll carry on...
    .
    Last edited by Wilmer; September 19th 2011 at 02:39 PM.
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  11. #11
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    Re: Triangle line lengths problem

    Thanks Wilmer

    I can see a glaring oversight! Line AB needs to have a length or line BD can cross line AE at any point! Apologies for the nonsense. So, AB=60.

    Using Cosine Law (which is new to me)

    BD = sqrt (50*50 + 60*60 - 2 * 50 * 60 cos(85))
    BD 74.67975324

    Do you agree?

    This leaves the lengths of lines DC and AC to calculate. Please can you point me towards the steps for these - I will then just check with you that the answers I calculate are correct and leave you in peace! Thanks for the early patience whilst i realised the omission!

    Regards

    Luke
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  12. #12
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    Re: Triangle line lengths problem

    Quote Originally Posted by LukeP20110907 View Post
    I can see a glaring oversight! Line AB needs to have a length or line BD can cross line AE at any point! Apologies for the nonsense. So, AB=60.
    You FINALLY realise that! Why do you think I told you: "OK, show me how YOU'd do that"?

    PLUS I told you that in this post (Post#5):
    > Well, had another look at your problem: only qs can be calculated.
    > r can be anywhere on the line that you show it on

    I'll get back to you later on lines DC and AC.

    T'a calculation pour BD est parfaite: good stuff!
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  13. #13
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    Re: Triangle line lengths problem

    Ok; so far we have calculated the following:
    angleADE = u = 65; angle DAB = v = 85
    lineAE = a = ~59.70; lineBD = b = ~74.68

    Now to calculate lineAC and lineCD:

    Step#1: using triangleABD, calculate angleABD = w:
    50 / SIN(w) = b / SIN(v)
    w = ASIN[50SIN(v) / b] = ~41.83

    Step#2: using triangleADE, calculate angleDAE = x:
    60^2 = 50^2 + a^2 - 2(50)(a)COS(x)
    x = ACOS[(50^2 + a^2 - 60^2) / (100a)] = ~65.62

    So angleBAC = y = v - x = ~19.38
    and angleACB = z = 180 - w - y = ~118.79

    Step#3: using triangleABC, calculate AC:
    AC / SIN(w) = 60 / SIN(z)
    AC = 60SIN(w) / SIN(z) = ~45.66

    Step#4: using triangleABC, calculate BC:
    BC / SIN(y) = 60 / SIN(z)
    BC = 60SIN(y) / SIN(z) = ~22.72

    So CD = BD - BC = 74.68 - 22.72 = ~51.96

    There you go...hope you can follow that...
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  14. #14
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    Re: Triangle line lengths problem

    Many, many thanks Wilmer

    I will be working through this today and putting into a spreadsheet.

    Thanks for the patience.

    Luke
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