Question

in triangle ABC , SIGMA a cosA =​

Answer 1

Given :   triangle ABC

To Find :  ∑acosA

Solution:

a/SinA = b/SinB = c/SinC = 2R

=> a = 2RSinA  , b = 2RsinB  , c = 2RsinC

∑acosA  = acosA + bcosB + ccosC

= 2RSinAcosA + 2RSinBcosB + 2RSinCcosC

sin2x = 2sinxcosx

sinC + sinD = 2sin(C + D)/2 Cos(C - D)/2

= R( Sin2A + Sin2B + 2SinCcosC)

= R ( 2Sin(A + B)cos(A - B)  + 2SinCcosC )

A + B =  π - C  => Sin(A+ B) = SinC

= R ( 2SinCcos(A - B)  + 2SinCcosC )

= 2RSinC ( Cos(A - B) + CosC)

CosC + CosD = 2Cos(C + D)/2Cos(C -D)/2

= 2RSinC 2 Cos{(A + C - B)/2} Cos{( A- B - C)/2}

=  2RSinC 2 Cos{(π-B - B)/2} Cos{( A- (π-A))/2}

= 2RSinC 2 Cos{(π-B - B)/2} Cos{( A- (π-A))/2}

= 4RSinC   Cos(π/2-B ) Cos ( A-π/2 )

= 4RSinC  SinB SinA

= 4RSinASinBSinC

= 4R ΠSinA

∑acosA  = 4R ΠSinA  or  4RSinASinBSinC

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