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Math Help - Finding the length of side in a 3d diagram with a few data only

  1. #1
    Newbie
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    Jan 2011
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    Finding the length of side in a 3d diagram with a few data only

    Hello guys,
    Here is the question:

    "The following three-dimensional diagram shows the four point A, B, C and D
    A, B and C are in the same horizontal plane and AD is vertical.
    Angle ABC is 45°
    BC = 50m
    Angle ABD = 30°
    and the last on is that Angle ACD is 20°

    Using the cosine rule in the trianlge ABC, or otherwise, find AD!"

    So the point D is directly above A.
    So you can imagine this as a pyramid with different long sides.
    So one triangle with D above A as it says.

    And now this is as far as I got:
    It says that AD is vertical so that we can say that the angles CAD and BAD must be 90 °

    Now we have the two triangles on the sides with only 2/3(we donīt need the third angle i think).

    To make all algebra easier I named BA as c
    and named CA as b
    and AD is X what we are looking for

    because we have now to rectangular triangles we can use tan twice.

    tan 30 = x/c
    and tan 20 = x/b

    or in the end: c tan 30 = b tan 20

    i looked back to the question and saw "use the cosine rule".
    The only combination which would make sense is this:

    bē=aē+cē-2ac cos (beta)
    or
    bē=50ē+cē-100c cos 45°

    We also could change: c tan 30 = b tan 20 to
    b=(c tan 30)/tan20 and square it

    bē= (cē x (tan 30)ē)/(tan 20)ē

    now we could "eliminate bē) couldnīt we to find c so therefore we have:

    (cē x (tan 30)ē)/(tan 20)ē = 50ē+cē-100c cos 45°

    Iīve tried several ways but I cant get to c in this equation because I always end up with a cē and a c but cant devide the c.
    I hope that is right what Iīve done and that someone can help me please.

    Thank you
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  2. #2
    Senior Member
    Joined
    Dec 2010
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    Rewrite the equation in the form A_1c^2 + A_2c + A_3 = 0 with A_i constants. You know how to solve a quadratic equation for c?
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  3. #3
    Super Member

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    Lexington, MA (USA)
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    Hello, ananas1301!

    You have a good game plan . . .


    \text{The diagram shows the four points  }A, B, C\text{ and }D.
    A, B\text{ and }C\text{ are in the same horizontal plane.}
    AD\text{ is vertical; }\,BC = 50\text{ m.}
    \angle ABC = 45^o;\;\angle ABD = 30^o;\;\angle ACD = 20^o.°

    \text{Using the cosine rule in }\Delta ABC\text{, or otherwise, find }AD.

    Let h = AD.
    Let a = BC = 50,\;b = AC,\;c = AB

    In right triangle D\!AC:\;\tan70^o \,=\,\dfrac{b}{h} \quad\Rightarrow\quad b \,=\,h\tan70^o

    In right triangle D\!AB:\;\tan60^o \,=\,\dfrac{c}{h} \quad\Rightarrow\quad c \,=\,h\tan60^o


    In \Delta ABC:\;b^2 \:=\:a^2 + c^2 - 2ac\cos B

    We have: . (h\tan70^o)^2 \;=\;50^2 + (h\tan60^o)^2 - 2(50)(h\tan60^o)\cos45^o

    . . . . . . . . h^2\tan^270^o \;=\;2500 + h^2\tan^260^o - 100h\tan60^o\cos45^o


    Since \tan60^o = \sqrt{3}\,\text{ and }\,\cos45^o = \frac{\sqrt{2}}{2}

    . . h^2\tan^270^o \;=\;2500 + h^2\cdot3 - 100h(\sqrt{3})\left(\frac{\sqrt{2}}{2}\right)

    . . (\tan^270^o-3)h^2  + 50\sqrt{6}\,h - 2500 \;=\;0


    Now apply the Quadratic Formula . . .

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  4. #4
    Newbie
    Joined
    Jan 2011
    Posts
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    Ohh yeah,

    now I saw it.

    I just didnīt recognize the quadratic formula.

    Now I put it in order and solved it.

    The book says 13.6m and I got 13.56.
    So its right.

    Thank you very much guys.
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