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![[Post New]](/templates/default/images/icon_minipost_new.gif) 23 Feb 2008 15:06:06 IST
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if the sides a,b and c of a triangle ABC with fixed area be changed by da,db and dc, show that dacosA+dbcosB+dccosC=0
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23 Feb 2008 16:43:59 IST
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d=0
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23 Feb 2008 22:36:12 IST
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a=2RsinA b=2RsinB c=2RsinC
da=2RcosAdA db=2RcosBdB dc=2RcosCdC
A+B+C =pi
=> dA + dB +dC =0
putting their values...
da/cosA + db/cosB +dc/cosC=0...
but how to prove
dacosA+dbcosB+dccosC=0??
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24 Feb 2008 07:51:16 IST
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How does R remain constant?
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Time wounds all heels |
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24 Feb 2008 11:49:22 IST
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umm...
i think it should be given in the question that R is constant...
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8 Mar 2008 21:23:17 IST
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hii
now i am assuming tat R is a constant ... coz it seems tat it is indeed needed ..
now area is already ocnstant .. (given)
so, area/abc = R ..
this means abc is constant ...
so, abc = k
or sinA.sinB.sinC = k
or sin2A + sin2B + sin2C = k
thus a.cosA + b.cosB + c.cosC = constant ..
now just equate now and after n get the answer ..
cheers
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Puneet Agrawal
IIT Delhi
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