Question

acosA+bcosB+ccosC=?​

Answer 1

Answer:

acosA+bcosB+ccosC=2asinBsinC

Step-by-step explanation:

From sine law,

a/sin A = b/sin B = c/sin C = k

⇒a= k sinA ,b= k sinB ,c= k sinC ........(i)

Now,

L.H.S.=acosA+bcosB+ccosC

=ksinAcosA+ksinBcosB+ksinCcosC

=k/2[2sinAcosA+2sinBcosB+2sinCcosC]

=k/2[sin2A+sin2B+sin2C]  

Now, consider any triangle ABC:

sin2A+sin2B+sin2C=4sinAsinBsinC......[form (i)]

⇒k/2[sin2A+sin2B+sin2C] = k/2[4sinAsinBsinC]

=2ksinAsinBsinC  

=2asinBsinC=R.H.S.