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tan(<a)=L/A
solving for L
L=A*tan(<a)
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[ This Message was edited by: kimcheeboi on 2004-10-17 05:39 ]
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Posted: 2004-10-17 06:36:00
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Yes. Kim is right. Because L/A = tg
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Posted: 2004-10-17 06:49:50
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Quote:tan(
solving for L
L=A*tan(
??? thx for tryin kim, but that explaination is a little harder than the first :s !
1) tan (2) how can the length of L = length of A x Tan(a)[tan a involves L neway, L cannot be on 2 sides of the equation?)
sorry my studpidness, but in AUS we do things just a tad different!
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Posted: 2004-10-18 00:08:50
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a is already known, and angle a is already known too.
so if you solve for a, it still works out.
Basically you're setting L/A=L/A, and sincve L/A=tan(a) L/A=tan(a)
kinda like chemistry and physics laws :-/
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Posted: 2004-10-18 00:17:00
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Quote:a is already known, and angle a is already known too.
so if you solve for a, it still works out.
Basically you're setting L/A=L/A, and sincve L/A=tan(a) L/A=tan(a)
kinda like chemistry and physics laws :-/
thx a lot, that cleared it up. would have been easy if i did chemistry and physics i guess!
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Posted: 2004-10-18 04:17:38
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no probs, man
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Posted: 2004-10-18 04:21:00
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OVer me head..way over me head
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Posted: 2004-10-18 15:09:33
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So, after having seen some computations and drawings here, I just now wonder how the bridge will look like?
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Posted: 2004-10-18 15:13:56
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Posted: 2004-10-18 17:01:58
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its just basic algebra.
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Posted: 2004-10-19 03:22:00
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