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Math Help - Goemetrically-derived formula for a new solid; need help solving for a dihedral angle

  1. #1
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    Post Goemetrically-derived formula for a new solid; need help solving for a dihedral angle

    Hello. So I don't know that anyone cares about where this problem comes from, but I'd be happy to explain: it's based on a new seven sided solid called the Chestahedron, which has all seven faces of equal area (four equilateral triangles and three 'kite' shapes). The formula below comes from setting the area of the unit equilateral triangles equal to the area of the kite as expressed in terms of the dihedral angle of the base equilateral triangle to the 'petal' equilateral triangles (the other 3).

    The long and the short of it is that I need help solving for the angle theta in the following equation:

    [IMG]file:///C:/Users/SETHMI%7E1/AppData/Local/Temp/moz-screenshot-1.png[/IMG]

    Any help would be appreciated.

    p.s. If this isn't solvable, then there's something wrong with the way the equation was derived, so help in that direction could be helpful as well.
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  2. #2
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    Hello arapacana

    Welcome to Math Help Forum!
    Quote Originally Posted by arapacana View Post
    Hello. So I don't know that anyone cares about where this problem comes from, but I'd be happy to explain: it's based on a new seven sided solid called the Chestahedron, which has all seven faces of equal area (four equilateral triangles and three 'kite' shapes). The formula below comes from setting the area of the unit equilateral triangles equal to the area of the kite as expressed in terms of the dihedral angle of the base equilateral triangle to the 'petal' equilateral triangles (the other 3).

    The long and the short of it is that I need help solving for the angle theta in the following equation:

    [IMG]file:///C:/Users/SETHMI%7E1/AppData/Local/Temp/moz-screenshot-1.png[/IMG]

    Any help would be appreciated.

    p.s. If this isn't solvable, then there's something wrong with the way the equation was derived, so help in that direction could be helpful as well.
    I think it's highly unlikely that there's an analytical solution to this equation. But, using Excel, I have found that, correct to 5 d.p., the first positive solution is
    \theta = 2.30532 radians \approx 132.085^o
    The next one lies between 4.2 and 4.3 radians; the next is around 8.6.

    Grandad
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