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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Apr 2008 21:31:38 IST
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If the bisector of angle A of triangle ABC makes an angle @ with BC, then sin@ is equal to .......
1.cos[(B-C)/2] 2.sin[(B-C)/2] 3.sin[B-(A/2)] 4.sin[C-(A/2)]
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3 Apr 2008 21:37:39 IST
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Hey, In triangle ABC, @=A/2 +C (ex ang) A/2= (Pi-B-C)/2 Therefore, @=((pi/2)-B/2-C/2) sin@= sin((pi/2)+ C/2-B/2) =cos((C-B)/2) sin@=cos( (B-C)/2) dunno why this is in blue
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GIVE LIFE THE BEST....srini...
DO OR DIE, OR BETTER NEVER TRY!!
I dont want to be the most intelligent among the ten.i pray the other nine RE fools!???!!!!!!
it doesn't matter if you win... whats important to me is i win!!!!!!!! |
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3 Apr 2008 21:38:29 IST
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is the ans (a)
solved it through substitution....
took a right angled isosceles triangle..........
@=pi/2
B+C=pi/4
sin@=1
only option a satisfies the given condition hence (a)......
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3 Apr 2008 21:39:57 IST
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bt ans given is (4)
dunno y
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4 Apr 2008 17:38:39 IST
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ne1 plz solve it
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4 Apr 2008 21:20:53 IST
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ne1 reply plz
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6 Apr 2008 21:42:09 IST
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ne1 plz solve it
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6 Apr 2008 21:59:33 IST
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no..it cant b 4th option..ans written by srini is rite...
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nobody is perfect..im nobody..datz y im here... |
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