inequality in triangle

harry654

apply (a²+b²)cos(α-β)>= a²+b²-h²

and then I dont know aaaaa :(

tiny-tim

no, you've missed out a cos(α-β)

harry654

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...

tiny-tim

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

harry654

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

tiny-tim

sorry, i'm not following you at all

harry654

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

harry654

How do you get (a²+b²)cos²(α-β) ≤ a²+b² - h² ?

harry654

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:(

tiny-tim

[QUOTE=harry654;3006244]I have a cosine law …/QUOTE]

what cosine law did you use?

harry654

I use this cosines law h²=a²+b²-2abcos(α-β) h is |DB| and then I dont know how carry on

tiny-tim

ok then 2ab = (a²+b²-h²)/cos(α-β) …

substitute that into the inequality

harry654

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

tiny-tim

uhh?

just write out the inequality, with 2ab replaced by (a²+b²-h²)/cos(α-β)

harry654

yes I see, but what inequality do you think ?

tiny-tim

what are you talking about?

there is only one inequality

harry654

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