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Question

In a triangle ABC,
a3cos(BC)+b3cos(CA)+c3cos(AB)=


A

abc

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B

3 abc

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C

a+b+c

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D

None of these

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Solution

The correct option is B

3 abc


a3cos(BC)+b3cos(CA)+c3cos(AB)=k3sin3Acos(BC)+k3sin3Bcos(CA)+k3sin3Ccos(AB)=12k3[sin2A(sin2B+sin2C)+sin2B(sin2C+sin2A)+sin2C(sin2A+sin2B)]=12k3[sin2A(2sinBcosB+2sinCcosC)+sin2B(2sinCcosC+2sinAcosA)+sin2C(2sinAcosA+2sinBcosB)]=k3[sinAsinB(sinAcosB+cosAsinB)+sinBsinC(sinBcosC+cosBsinC)+sinCsinA(sinCcosA+cosCsinA)]=k3[sinAsinB(sin(A+B))+sinBsinC(sin(B+C))+sinCsinA(sin(C+A))]=k3[sinAsinB(sin(πC))+sinBsinC(sin(πA))+sinCsinA(sin(πB))]=k3[sinAsinBsinC+sinBsinCsinA+sinCsinAsinB]=3k3sinAsinBsinC=3abc.


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