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Thread: How to find the altitude and size of an object atop a building

  1. #1
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    How to find the altitude and size of an object atop a building

    This is my first post here so please bear with me as this is probably a well known problem with an easy solution that I most likely should know at age 28, however, seeing as how I don't, I thought I would try asking people who are smarter than me for help. I am trying to figure this one out without a long formula. Probably doesn't exist, but need an expert's and an experienced mathmatician's help first of all, to show me the standard formula in solving this problem as well as (if any) some tips and tricks or "unorthodox" so to speak ways in solving it.

    So, Dude and I, the two observers, are standing 1 mile apart from eachother.
    We both see in the distance a really tall building. On the very top of this building there is a huge Radome antenna.
    Neither of us know the height of the building nor do we know the size of Radome on top of the building.
    Using an astrolabe, we find the degrees of altitude of the Radome from both standpoints.
    Mine shows the Radome at 60° altitude and Dude's astrolabe shows 70° altitude.
    With these factors being the only measurements we have (we don't know the size of the radome, height of building)
    what is the mathematical formula(s) or equation(s) to find both the size of the Radome as well as how high off the ground it is.
    (The building in this problem is technically irrelevant considering we don't know the height of it)

    Again, I'm probably going to get some responses saying this is so easy a 5th grader could figure it out, basically, if not saying it bluntly, that I'm a moron. But I realize this going into it and my whole goal in asking this question is to learn how to solve the problem because I don't know how. Or I just don't remember what my teacher taught me over 10 years ago. Please refresh my memory or teach me the right way and hopefully some short cut way as well. Thanks!2234
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    I used to be good at math until they decided to mix in the greek alphabet.

    2154

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    Quote Originally Posted by missizfizziks View Post
    This is my first post here so please bear with me as this is probably a well known problem with an easy solution that I most likely should know at age 28, however, seeing as how I don't, I thought I would try asking people who are smarter than me for help. I am trying to figure this one out without a long formula. Probably doesn't exist, but need an expert's and an experienced mathmatician's help first of all, to show me the standard formula in solving this problem as well as (if any) some tips and tricks or "unorthodox" so to speak ways in solving it.

    So, Dude and I, the two observers, are standing 1 mile apart from each other.
    We both see in the distance a really tall building. On the very top of this building there is a huge Radome antenna.
    Neither of us know the height of the building nor do we know the size of Radome on top of the building.
    Using an astrolabe, we find the degrees of altitude of the Radome from both standpoints.
    Mine shows the Radome at 60° altitude and Dude's astrolabe shows 70° altitude.
    With these factors being the only measurements we have (we don't know the size of the radome, height of building)
    what is the mathematical formula(s) or equation(s) to find both the size of the Radome as well as how high off the ground it is.
    (The building in this problem is technically irrelevant considering we don't know the height of it)

    Again, I'm probably going to get some responses saying this is so easy a 5th grader could figure it out, basically, if not saying it bluntly, that I'm a moron. But I realize this going into it and my whole goal in asking this question is to learn how to solve the problem because I don't know how. Or I just don't remember what my teacher taught me over 10 years ago. Please refresh my memory or teach me the right way and hopefully some short cut way as well. Thanks!2234
    We need some definitions and assumptions:

    Define the height of the building?

    Define the size of the Radome?

    Are we to assume that you and dude are using astrolabe of equal height and that height is negligible compared to the size of the Radome and height of the building?
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #3
    The scale in the picture is incorrect, which can be misleading. The person on the left would certainly be further from the building.

    I would first begin by partitioning the picture into two triangles, the left and the right. Next, we should label the base lengths of both triangles. If we label the base length of the left triangle as x, then we must label the right triangle's base length as 1-x.

    We would have to use some basic trig. next, but people here would like to see your work first and then answer any questions.

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    Quote Originally Posted by Denis View Post
    Your problem seems to be this simple:
    Code:
               C
    
    
    
    
    
    A          D        B
    Given: triangle ABC, AB = 1, angleCAB = 60, angleCBA = 70.
    CD is height line. Calculate length of CD.

    So we know that angleACB = 50; also angleACD = 30, angleBCD = 20.

    CD can easily be calculated:
    use Law of Sines to calculate AC
    ACD being a 30-60-90 triangle, then AD = AC/2.

    Use google to get familiar with Law of Sines, and the 30-60-90 triangle.

    If you also need the height of building and antenna,
    then one more information is required: width of building will do.
    I don't see how the width of the building is relevant here!
    According to the given drawing, one of the lines of site (BC - according to your nomenclature), is not touching the building.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

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    Quote Originally Posted by Denis View Post
    Yes, saw that...I assume it's just bad drawing...
    else not much makes sense
    As you implied, depending on the height of the Radome and assuming the 60[tex]^{\circ}[/tex] is correct, you might not be able to see the Radome from the 70[tex]^{\circ}[/tex] position. Also, you need to assume that the astrolabe is at ground level, the ground is flat, etc. or just assume (there's that word again) that the difference is buried in round-off.

    If you swap the angles with each other, the drawing makes sense (but is certainly not to scale) and does not appreciably change the problem so that your solution given is [with a couple of changes of exact values] essentially correct and your comment about the width of the building is then sufficient.


    EDIT:
    Last edited by Ishuda; 06-30-2016 at 11:18 AM.

    -Ishuda

  6. #6
    Just realized that OP wants to know the height of the antenna... don't think there is enough information.

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    Cool

    Quote Originally Posted by ecoo View Post
    Just realized that OP wants to know the height of the antenna... don't think there is enough information.
    One can find the height of the top of the antenna but, no, nothing else can be determined.

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    Quote Originally Posted by stapel View Post
    One can find the height of the top of the antenna but, no, nothing else can be determined.
    the height of the top of the antenna = 1.06233743 miles !!!!
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  9. #9
    Quote Originally Posted by stapel View Post
    One can find the height of the top of the antenna but, no, nothing else can be determined.
    Darn English, I should have said length of the antenna.
    Last edited by ecoo; 06-30-2016 at 12:17 PM.

  10. #10
    Elite Member stapel's Avatar
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    Quote Originally Posted by Subhotosh Khan View Post
    the height of the top of the antenna = 1.06233743 miles !!!!
    I'm getting something similar. For clarity, I'm using the following derived diagram:

    Code:
    diagram:
             *
            /|\
           / | \
         /  h|  \
        /    |   \
      /      |    \
     /60     |   70\ 
    *--------*------*
    |<--y--->|<---->|
              5280-y
    . . . . .[tex]\dfrac{h}{y}\, =\, \tan(60^{\circ})[/tex]

    . . . . .[tex]\dfrac{h}{5,280\, -\, y}\, =\, \tan(70^{\circ})[/tex]

    Solving for "y=":

    . . . . .[tex]\dfrac{h}{\tan(60^{\circ})}\, =\, y[/tex]

    . . . . .[tex]\dfrac{h}{\tan(70^{\circ})}\, =\, 5,280\, -\, y[/tex]

    . . . . .[tex]\dfrac{h}{\tan(70^{\circ})}\, -\,5,280\, =\, -y[/tex]

    . . . . .[tex]y\, =\, 5,280\, -\, \dfrac{h}{\tan(70^{\circ})}[/tex]

    This allows me to form an equation only in "h", which is the only variable I care about anyway:

    . . . . .[tex]5,280\, -\, \dfrac{h}{\tan(70^{\circ})}\, =\, \dfrac{h}{\tan(60^{\circ})}[/tex]

    . . . . .[tex]5,280\, =\, h\, \left(\dfrac{1}{\tan(60^{\circ})}\, +\, \dfrac{1}{\tan(70^{\circ})}\right)[/tex]

    . . . . .[tex]\dfrac{5,280}{\left(\dfrac{1}{\tan(60^{\circ})}\, +\, \dfrac{1}{\tan(70^{\circ})}\right)}\,=\, h[/tex]

    So the result is an utterly unreasonable value.

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