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Thread: help with a formula to find lengths

  1. #1

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    help with a formula to find lengths

    Hi, need some help to figure out a formula to find the following(check attached)
    Given:
    AD=3000, AB=6000 and BC=800
    AL=LM=MB=2000
    Find:
    LP, MN, DP, PN, NC and the degrees for ADP corner

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    Thanks in advance

  2. #2
    Only Slightly Obsessive jemidiah's Avatar
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    Re: help with a formula to find lengths

    I'll assume angles that look like right angles are right angles.

    Drop a perpendicular from C to the line AD, and call the point of intersection Y. This line intersects LP at one point, say Q, and it intersects MN at one point, say R. Evidently AY = LQ = MR = BC = 800. It follows that since AB was cut into thirds, DC was as well, so DP = PN = NC = DC/3. As for DC, the triangle CYD is a right triangle, so we use the Pythagorean theorem to find CD. The leg CY has length BA = 6000. The leg YD has length AD - AY = AD - BC = 2200. Hence DC = Sqrt(6000^2 + 2200^2) ~= 6391. In particular DP=PN=NC ~= 2130. The angle ADP has tangent equal to the opposite leg CY divided by the adjacent leg DY, hence the angle is arctan(CY/DY) = arctan(6000/2200) ~= 70 degrees.

    Now LP = LQ + QP and MN = MR + RN, so we need QP and RN. For that, use similar triangles. In particular the triangle CYD is similar to the triangles CQP and CRN. We have CY/YD = CQ/QP = CR/RN, so QP = YD*CQ/CY and RN = YD*CR/CY. Hence QP = 2200*(2AB/3)/6000 = 2200*4000/6000 ~= 1467 and RN = 2200*(AB/3)/6000 ~= 733.
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  3. #3

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    Re: help with a formula to find lengths

    Thanks for your reply.. i was actually able to work it out this way. but i was looking for a simplified formula to be passed as a parameter to a program i.e.
    MN=BC+((AD-BC)/3)
    LP=BC+(2*(AD-BC)/3)
    CN=sqrt( [((AD-BC)/3)]^2 + MB^2)

    not sure if this will always work !! and still need the ADP corner

  4. #4
    New Member henrytrs35's Avatar
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    Re: help with a formula to find lengths

    Draw line SC perpendicular to AD passing thru C. You now have a triangle SDC. You know the lengths of DP, PN and NC, right? Sum it and you have the length of DC. AB is just AL + LM + MB. AB = SC. Last leg, SD = DA - CB. DA and CB are both given. So you already know the length of all the sides of the triangle SDC. You can get ADP by applying sine law using the right angle and ADP as the unknown. Hope it helps(and is correct).

  5. #5
    Only Slightly Obsessive jemidiah's Avatar
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    Re: help with a formula to find lengths

    Your reply confuses me in several ways. (1) I showed you how to find the ADP corner, but you seem to have ignored this. (2) If you could do what I did in my response, why bother asking the question in the first place? The specific numbers don't matter so long as "the picture is true", eg. the angles that look like right angles are right angles, the base AD is larger than BC, that sort of thing. (3) Why not post your formulas first and ask for confirmation if that's what you were really after? Indeed, why post numbers at all if you're just interested in the formulas?

    In any case, your formulas are correct. The angle formula implied by my post is
    angle ADP = arctan(AB/(AD-BC))

    Note that inverse trig functions frequently return radians, depending on which calculator/programming language you use, so you may have to convert to degrees.
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