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Inequality in triangle

by harry654
Tags: inequality, triangle
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tiny-tim
#19
Nov27-10, 09:02 AM
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Quote Quote by harry654 View Post
apply (a+b)cos(α-β)>= a+b-h
no, you've missed out a cos(α-β)
harry654
#20
Nov27-10, 09:09 AM
P: 58
Quote Quote by tiny-tim View Post
no, you've missed out a cos(α-β)
2ab=(a+b-h)/cos(α-β) Is that OK?

then a+b>=2ab
a+b>=(a+b-h)/cos(α-β) /*cos(α-β)
(a+b)cos(α-β) ? (a+b-h)
so? I am lost...
tiny-tim
#21
Nov27-10, 09:24 AM
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i can't see where you got lost

you should have arrived at (a+b)cos(α-β) ≤ a+b - h
carry on from there
harry654
#22
Nov27-10, 09:35 AM
P: 58
you should have arrived at (a+b)cos(α-β) ≤ a+b - h ( but why is there ≤ and not >=)

(a+b)cos(α-β) ≤ a+b - h


(a+b)cos(α-β) ≤ 2ab
tiny-tim
#23
Nov27-10, 09:43 AM
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Quote Quote by harry654 View Post
(a+b)cos(α-β) ≤ a+b - h


(a+b)cos(α-β) ≤ 2ab
sorry, i'm not following you at all
harry654
#24
Nov27-10, 09:48 AM
P: 58
thank you for your patience
I think I never solve this problem
harry654
#25
Nov27-10, 10:45 AM
P: 58
How do you get (a+b)cos(α-β) ≤ a+b - h ?
harry654
#26
Nov27-10, 12:40 PM
P: 58
I really dont know how carry on...
I have a cosine law and then I dont know how get (a+b)cos(α-β) ≤ a+b - h
Please help me:(
tiny-tim
#27
Nov27-10, 01:35 PM
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[QUOTE=harry654;3006244]I have a cosine law /QUOTE]

what cosine law did you use?
harry654
#28
Nov27-10, 01:55 PM
P: 58
I use this cosines law h=a+b-2abcos(α-β) h is |DB| and then I dont know how carry on
tiny-tim
#29
Nov27-10, 02:06 PM
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Quote Quote by harry654 View Post
I use this cosines law h=a+b-2abcos(α-β) h is |DB|
ok then 2ab = (a+b-h)/cos(α-β)

substitute that into the inequality
harry654
#30
Nov27-10, 02:12 PM
P: 58
so and now the problem begins because I dont know how :(
tiny-tim
#31
Nov27-10, 02:25 PM
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uhh?

just write out the inequality, with 2ab replaced by (a+b-h)/cos(α-β)
harry654
#32
Nov27-10, 02:30 PM
P: 58
yes I see, but what inequality do you think ?
tiny-tim
#33
Nov27-10, 02:36 PM
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Quote Quote by harry654 View Post
yes I see, but what inequality do you think ?
what are you talking about?

there is only one inequality
harry654
#34
Nov27-10, 02:47 PM
P: 58
but I cant
in (a+b)cos(α-β)≤2ab
substitude 2ab because I must prove that inequality and when I substitude 2ab it isnt mathematical proof
tiny-tim
#35
Nov27-10, 02:53 PM
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do it now to see where you're going
you can tidy it up later!
harry654
#36
Nov27-10, 02:58 PM
P: 58
OK
I have (a+b)cos(α-β) ≤ (a+b-h)/cos(α-β)
and from it I get
(a+b)cos(α-β) ≤ a+b-h
but how can I tidy up later?


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