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Math Help - Geometry problem with triangles

  1. #1
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    Geometry problem with triangles

    How frustrating, I think when I was 11 years old I could have done this in approximately 30 seconds, now more than a quarter of a century later my brute force methods lead me to a horrific looking equation that can't be optimal!. Please check attached image, this is the projection of an image at one distance (100cm) viewed at another distance (160cm), I need to find out the "measured" size either side of the centre of the projection based on a presumed "tilt" at the second measuring surface. I am sure there is some kind of triangle based wizardry that makes this a simple matter to express the (unequal) two distances as a ratio to their (equal) values in an untilted world. Thanks for any help
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  2. #2
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    Quote Originally Posted by lsbi4 View Post
    I am sure there is some kind of triangle based wizardry that makes this a simple matter to express the (unequal) two distances as a ratio to their (equal) values in an untilted world.


    True enough there is. It is called the "Law of Sines", which says \frac{sin(A)}{a}=\frac{sin(B)}{b}=\frac{sin(C)}{c} (where the capital letters are angles and the lower-case letters are the lengths of the sides opposite the angles).

    For example, \frac{GX}{CX}=\frac{sin(<CGX)}{sin(<GCX)}.

    Hope this helps. Let me know if you need more info.
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  3. #3
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    Thanks, maybe I need more help, I only know the angle GXC, that is the one I am varying to see how it affects the ratio GX/XC. i.e. I don't know what angles GCX and CGX are
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  4. #4
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    well, playing with adding angles of triangles up, am i right in thinking that the angles are

    90 - tilt angle (BXT) + divergence angle (the half angle at the very top triangle)

    and

    90 + tilt angle - divergence angle

    but I suppose the divergence angle will vary based on the width of the triangles at the depths 100 and 160cm

    ohhhhhhh..............
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  5. #5
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    For example, if the distances

    DE = 5cm, XB = 8cm, then DAE is tan-1 of 1/20, or 2.86 degrees

    I think that means

    XT/XB = sin(90+2.86)/sin(90-2.86-tiltangleBXT)
    GX/CX = sin(90-tilt+2.86)/sin(90+tilt-2.86)

    or am I wrong?
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  6. #6
    Member pflo's Avatar
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    Quote Originally Posted by lsbi4 View Post
    I only know the angle GXC, that is the one I am varying to see how it affects the ratio GX/XC.
    It is possible to derive a formula for the required ratio based on that angle, but it is MUCH easier if you know something else as well. For example, do you know the measure of angle FAD?

    Also, will AD and AX always be the 100 and 160 respectively, or should these distances be treated as variables?
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  7. #7
    Member pflo's Avatar
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    Ok, I've worked on this for a fair bit now - I like puzzles. But I can't figure it out without knowing something else besides the lengths of AD and AX. Maybe you shouldn't be quite so frustrated!

    In general it is true - you've got to know three things about a triangle to know everything about it. The problem here is that the vertex angle (at A) also effects the ratio between Gx and CX.

    I am beginning to believe I was wrong in my previous post about the possibilities of deriving a formula for you. But again, it would be a fairly easy proposition if you know one more thing about this situation.
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  8. #8
    Member pflo's Avatar
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    Quote Originally Posted by lsbi4 View Post
    if the distances DE = 5cm, XB = 8cm,
    then DAE is tan-1 of 1/20, or 2.86 degrees
    I don't see how this works.
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  9. #9
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    Yeah, I just mean the angle DAE is the inverse tan of 1/20, in the case that DE = 5cm (a common real example of what it might measure in my situation), of course you don't need to include the fact that XB = 8cm, this just shows tan-1 of DAE is 8/160 rather than 5/100, i.e. the same thing. Of course specifying DE = 5cm is not a generic example!
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