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Inequality in triangle 
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#20
Nov2710, 09:09 AM

P: 58

then a²+b²>=2ab a²+b²>=(a²+b²h²)/cos(αβ) /*cos(αβ) (a²+b²)cos(αβ) ? (a²+b²h²) so? I am lost... 


#21
Nov2710, 09:24 AM

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i can't see where you got lost …
you should have arrived at (a²+b²)cos²(αβ) ≤ a²+b²  h² … carry on from there 


#22
Nov2710, 09:35 AM

P: 58

you should have arrived at (a²+b²)cos²(αβ) ≤ a²+b²  h² … ( but why is there ≤ and not >=)
(a²+b²)cos²(αβ) ≤ a²+b²  h² (a²+b²)cos(αβ) ≤ 2ab 


#24
Nov2710, 09:48 AM

P: 58

thank you for your patience
I think I never solve this problem 


#25
Nov2710, 10:45 AM

P: 58

How do you get (a²+b²)cos²(αβ) ≤ a²+b²  h² ?



#26
Nov2710, 12:40 PM

P: 58

I really dont know how carry on...
I have a cosine law and then I dont know how get (a²+b²)cos²(αβ) ≤ a²+b²  h² Please help me:( 


#27
Nov2710, 01:35 PM

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[QUOTE=harry654;3006244]I have a cosine law …/QUOTE]
what cosine law did you use? 


#28
Nov2710, 01:55 PM

P: 58

I use this cosines law h²=a²+b²2abcos(αβ) h is DB and then I dont know how carry on



#29
Nov2710, 02:06 PM

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substitute that into the inequality 


#30
Nov2710, 02:12 PM

P: 58

so and now the problem begins because I dont know how :(



#31
Nov2710, 02:25 PM

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uhh?
just write out the inequality, with 2ab replaced by (a²+b²h²)/cos(αβ) 


#32
Nov2710, 02:30 PM

P: 58

yes I see, but what inequality do you think ?



#33
Nov2710, 02:36 PM

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there is only one inequality 


#34
Nov2710, 02:47 PM

P: 58

but I cant
in (a²+b²)cos(αβ)≤2ab substitude 2ab because I must prove that inequality and when I substitude 2ab it isnt mathematical proof 


#35
Nov2710, 02:53 PM

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do it now to see where you're going …
you can tidy it up later! 


#36
Nov2710, 02:58 PM

P: 58

OK
I have (a²+b²)cos(αβ) ≤ (a²+b²h²)/cos(αβ) and from it I get (a²+b²)cos²(αβ) ≤ a²+b²h² but how can I tidy up later? 


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