inequality in triangle

by harry654
Tags: inequality, triangle
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P: 25,514
 Quote by harry654 apply (a²+b²)cos(α-β)>= a²+b²-h²
no, you've missed out a cos(α-β)
P: 58
 Quote by tiny-tim no, you've missed out a cos(α-β)
2ab=(a²+b²-h²)/cos(α-β) Is that OK?

then a²+b²>=2ab
a²+b²>=(a²+b²-h²)/cos(α-β) /*cos(α-β)
(a²+b²)cos(α-β) ? (a²+b²-h²)
so? I am lost...
 PF Patron HW Helper Sci Advisor Thanks P: 25,514 i can't see where you got lost … you should have arrived at (a²+b²)cos²(α-β) ≤ a²+b² - h² … carry on from there
 P: 58 you should have arrived at (a²+b²)cos²(α-β) ≤ a²+b² - h² … ( but why is there ≤ and not >=) (a²+b²)cos²(α-β) ≤ a²+b² - h² (a²+b²)cos(α-β) ≤ 2ab
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P: 25,514
 Quote by harry654 (a²+b²)cos²(α-β) ≤ a²+b² - h² (a²+b²)cos(α-β) ≤ 2ab
sorry, i'm not following you at all
 P: 58 thank you for your patience I think I never solve this problem
 P: 58 How do you get (a²+b²)cos²(α-β) ≤ a²+b² - h² ?
 P: 58 I really dont know how carry on... I have a cosine law and then I dont know how get (a²+b²)cos²(α-β) ≤ a²+b² - h² Please help me:(
 PF Patron HW Helper Sci Advisor Thanks P: 25,514 [QUOTE=harry654;3006244]I have a cosine law …/QUOTE] what cosine law did you use?
 P: 58 I use this cosines law h²=a²+b²-2abcos(α-β) h is |DB| and then I dont know how carry on
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P: 25,514
 Quote by harry654 I use this cosines law h²=a²+b²-2abcos(α-β) h is |DB|
ok then 2ab = (a²+b²-h²)/cos(α-β) …

substitute that into the inequality
 P: 58 so and now the problem begins because I dont know how :(
 PF Patron HW Helper Sci Advisor Thanks P: 25,514 uhh? just write out the inequality, with 2ab replaced by (a²+b²-h²)/cos(α-β)
 P: 58 yes I see, but what inequality do you think ?
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