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Question

In ABC, sides opposite to angles A,B,C are denoted by a,b,c respectively. Then the value of acosA+bcosB+ccosC=

A
2abcsinAsinBsinC
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B
2asinBsinC
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C
2csinAsinB
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D
2bsinCsinA
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Solution

The correct option is D 2bsinCsinA
Using sine rule, we have : asinA=bsinB=csinC=2R
We have,
acosA+bcosB+ccosC=2RsinAcosA+2RsinBcosB+2RsinCcosC=R(sin2A+sin2B+sin2C)=R×(4sinAsinBsinC)(sin2A+sin2B+sin2C=4sinAsinBsinC)=2(2RsinA)sinBsinC
=2asinBsinC
Similarly
acosA+bcosB+ccosC=2bsinCsinA
Similarly
acosA+bcosB+ccosC=2csinAsinB

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