Results 1 to 4 of 4

Math Help - Please Help, Horizontal Lengths!!!

  1. #1
    Junior Member
    Joined
    Nov 2008
    Posts
    41

    Unhappy Please Help, Horizontal Lengths!!!

    Hi all,

    I'm new here so big hi to everyone.

    O.k. I have an assignment question here, its the last of 5 and i'm chuffed because i've solved all the rest but now my heads a bit broken.

    this is the last question:

    "three lines on a site form a triangle on plan. the HORIZONTAL lengths were measured and found to be 20.988, 38.011 and 25.101.

    Determine the angles etc."

    I can answer the question no prob, but i'm being cautious here, the words horizontal indicates that these are not the lengths of the lines of the trianlge. they may well be and i'm just being paranoid but this seems a trick question to me.

    so my question is, can i find the length of an angled line from the horizontal length alone?

    I'm sure its impossible because i have no angles, with the angle its easy, additionally, i have no vertical component in order to find the angles using tan..........im lost.

    so guys, any thoughts or am i just reading to much into the question?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Nov 2008
    Posts
    82
    Quote Originally Posted by Parton-Bill View Post
    Hi all,

    I'm new here so big hi to everyone.

    O.k. I have an assignment question here, its the last of 5 and i'm chuffed because i've solved all the rest but now my heads a bit broken.

    this is the last question:

    "three lines on a site form a triangle on plan. the HORIZONTAL lengths were measured and found to be 20.988, 38.011 and 25.101.

    Determine the angles etc."

    I can answer the question no prob, but i'm being cautious here, the words horizontal indicates that these are not the lengths of the lines of the trianlge. they may well be and i'm just being paranoid but this seems a trick question to me.

    so my question is, can i find the length of an angled line from the horizontal length alone?

    I'm sure its impossible because i have no angles, with the angle its easy, additionally, i have no vertical component in order to find the angles using tan..........im lost.

    so guys, any thoughts or am i just reading to much into the question?

    Yes a definate trick question has been posed,
    I cant see a feasible way of computing the angles of the triangle unless the height and or classification of the triangle is known,

    if the horizontal lengths simply represent the actual lengths of the triangle and by imposing that we know nothing about the shape of the triangle and thus assume to be scalene then to solve the angles, simply employ the cosine rule.

    Hope this helps,

    David
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2008
    Posts
    41
    Thanks David,

    thanks very much for that matey, i came to the same conclusion.

    after trying everything else i still feel its a trick question but with no other information Ive decided to use cosine.

    many thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,740
    Thanks
    645
    Hello, Parton-Bill!

    The term "horizontal" was misused (and highly misleading) here.
    I think those three lengths are simply the sides of the triangle. .**


    A triangle has sides: . a= 20.988,\;\;b= 38.011,\;\;c= 25.101

    Determine the angles of the triangle.
    Code:
                      B
                      *
                    *  *
         25.101   *     * 20.998
                *        *
              *           *
            *              *
        A *  *  *  *  *  *  * C
                38.011
    Use the Law of Cosines . . .


    \cos A \;=\;\frac{b^2+c^2-a^2}{2bc} \;=\;\frac{38.001^2 + 25.101^2-20.998^2}{2(38.011)(25.101)} \;=\;0.856281392 \quad\Rightarrow . . A \;\approx\;31.1^o

    \cos B \;=\;\frac{a^2+c^2-b^2}{2ac} \;=\;\frac{20.998^2 + 25.101^2-38.011^2}{2(20.998)(25.101)} \;=\;\text{-}0.354658159 \quad\Rightarrow . B \;\approx\;110.8^o

    \cos C \;=\;\frac{a^2+b^2-c^2}{2ab} \;=\;\frac{20.998^2 + 38.011^2 - 25.101^2}{2(20.998)(38.011)} \;=\;0.78662162 \quad\Rightarrow . . . C \;\approx\;38.1^o


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **

    If those were truly horizontal distances,
    . . then \Delta ABC is "projected" onto a horizontal line.
    Code:
                    B
                    *
                   *: *
                  * :   *
                 *  :     *
                *   :       * C
               *    :   *   :
              *     *       :
             *  *   :       :
          A *       :       :
            :       :       :
        - - * - - - * - - - * - -
            A'      B'      C'

    And we see that: . A'B' + B'C\:\!'\:\text{ must equal }\:A'C\:\!'
    . . . which is not true of the given segments.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Lengths of Curves
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 3rd 2010, 05:46 PM
  2. Horizontal Angle into Horizontal Circle
    Posted in the Calculus Forum
    Replies: 0
    Last Post: March 22nd 2010, 11:18 AM
  3. Arc Lengths
    Posted in the Calculus Forum
    Replies: 6
    Last Post: October 22nd 2009, 01:59 AM
  4. Curve lengths...
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 29th 2008, 04:37 PM
  5. Arc Lengths
    Posted in the Calculus Forum
    Replies: 0
    Last Post: September 17th 2007, 02:09 PM

Search Tags


/mathhelpforum @mathhelpforum