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I am building a wood model of a Icosahedron.
Can anybody tell me how I work out the angle that any two ajacent faces meet?
If you could direct me to formula for this sort of thing I would be happy to work it out for myself.
SoapyJoe
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138.19 degrees:)
Last edited by liuv (2006-05-19 19:58:06)
I'm from Beijing China.
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Thanks Liuv,
Can you please tell me how you did this?
Soapy
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O-ABCDE is a part of a Icosahedron.
OA=OB=OC=OD=OE=AB=BC=CD=DE=EA
Since ABCDE is a Regular Pentagon,therefore <EAB=108°.
triangle OAB, OBC, OCD, ODE, OEF are Regular triangles.
let OA=OB=OC=OD=OE=AB=BC=CD=DE=EA=1,
OF=AF=0.5
then
BF=EF=sqrt(AB^2-AF^2)=sqrt(3)/2
in triangle EAB
BE^2=EA^2+AB^2-2*EA*AB*Cos<EAB=2-2Cos108°
in triangle EFB
BE^2=EF^2+FB^2-2*EF*FB*Cos<EFB=3/2-3/2*Cos<EFB
therefore
2-2Cos108°=3/2-3/2*Cos<EFB
<EFB=138.19°
Last edited by liuv (2006-05-19 23:16:11)
I'm from Beijing China.
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I am building a wood model of a Icosahedron.
Can anybody tell me how I work out the angle that any two ajacent faces meet?
If you could direct me to formula for this sort of thing I would be happy to work it out for myself.
SoapyJoe
Hi joe! We're you ever able to figure out the adjacent angles? I'm building an icosahedron out of one inch thick wood, so I wanted to know what angles to cut the adjacent edges. Nt help would be greatly appreciated!
hi Evan,
Welcome to the forum.
I had to search back but I remembered this has come up in another thread.
Have a look at
http://www.mathisfunforum.com/viewtopic.php?id=16061
post 30 if you want the method or
http://en.wikipedia.org/wiki/Table_of_p … ral_angles
if you just want to look them up.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Thanks, Bob!
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