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Question

In a ΔABC, bcos(C+θ)+ccos(Bθ)=


A
asinθ
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B
acosθ
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C
atanθ
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D
acotθ
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Solution

The correct option is B.acosθ

In ABC,A+B+C=π...(i) and

a=2RsinA,b=2RsinB,c=2RsinC...(ii)

Now,

bcos(C+θ)+ccos(Bθ)=R[2sinBcos(C+θ)+2sinCcos(Bθ)] [From (ii)]

=R[sin(B+C+θ)+sin(BCθ)+sin(B+Cθ)+sin(CB+θ)] [Using 2sinxsiny=sin(x+y)+sin(xy)]

=R[sin(B+C+θ)+sin(B+Cθ)+sin(BCθ)+sin(CB+θ)]

=R[sin(π(Aθ))+sin(π(A+θ))+sin(BCθ)sin(BCθ)] [From(i)]

=R[sin(Aθ)+sin(A+θ)]

=R[2sinAcosθ]

=(2RsinA)cosθ

=acosθ

Thus, bcos(C+θ)+ccos(Bθ)=acosθ


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