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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: Properties Of Triangles
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[Post New]28 Mar 2008 08:49:23 IST
    Subject: Properties Of Triangles
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m.viddya (101)

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Q.In triangle ABC if cos 3A+cos3B+cos3C=1 then the angle is?
 
1.Acute
2.Obtuse
3.Equilateral
4.Such a triangle does not exist.
    
[Post New]28 Mar 2008 21:34:13 IST
    Subject: Re:Properties Of Triangles
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deepesh28 (12)

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COS3B+COS3C+COS3A=1
COS3B+COS3C+COS3A-1=0
COS3(B+C)/2.COS3(B-C)/2-SIN^3A/2=0
SOLVING WE GET
COS3C/2.COS3B/2.COS3A/2=0
EITHER3C/2=
C=/3     THEREFORE TRIANGLE IS ACUTE ANGLED
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[Post New]28 Mar 2008 22:38:53 IST
    Subject: Re:Properties Of Triangles
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sandeepramesh (1090)

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excuse me but why shd 3C/2 = pi?? I believe that that step is wrong :)
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[Post New]29 Mar 2008 06:43:33 IST
    Subject: Re:Properties Of Triangles
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eistien (343)

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hey if 3C/2=pie then isnt C=2(pie)/3 and not pie/3!!!!
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[Post New]29 Mar 2008 06:48:54 IST
    Subject: Re:Properties Of Triangles
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eistien (343)

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hey your method is perfectly right but 3C/2=pie/2 so C=pie/3
the same method i have tried
hey but just because one angle is pie/3 it does not mean it is acute it can be obtuse too!!!
 
COS3B+COS3C+COS3A=1
COS3B+COS3C+COS3A-1=0
COS3(B+C)/2.COS3(B-C)/2-SIN^3A/2=0
SOLVING WE GET
COS3C/2.COS3B/2.COS3A/2=0
 
which implies anyone of the 3 cos functions has to be zero
any one of the angles has to be pie/2
 
so sloving we get
C or A or B =pie/3
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