in triangle ABC , SIGMA a cosA =
Answers
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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