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In-a-ABC-a-2-sin-2-B-sin-2-C-

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Question Number 113848 by deepraj123 last updated on 15/Sep/20
In a △ABC, Σ a^2 (sin^2 B −sin^2 C) =
Ina△ABC,Σa2(sin2B−sin2C)=
Answered by 1549442205PVT last updated on 16/Sep/20
We have Σa^2 (sin^2 B−sin^2 C)= a^2 (sin^2 B−sin^2 C )+b^2 (sin^2 C−sin^2 A) +c^2 (sin^2 A−sin^2 B) From sine theorem we have: (a/(sinA))=(b/(sinB))=(c/(sinC))=2R⇒a^2 =4R^2 sin^2 A b^2 =4R^2 sin^2 B,c^2 =4R^2 sim^2 C.Replace into above equality we get Σa^2 (sin^2 B−sin^2 C) =4R^2 [sin^2 A(sin^2 B−sin^2 C)+sin^2 B(sin^2 C−sin^2 A) +sin^2 C(sin^2 A−sin^2 B)]. (Put x=sin^2 A,y=sin^2 B,z=sin^2 C ) =4R^2 [x(y−z)+y(z−x)+z(x−y)] =4R^2 (xy−xz+yz−yx+zx−zy)=0
WehaveΣa2(sin2B−sin2C)=a2(sin2B−sin2C)+b2(sin2C−sin2A)+c2(sin2A−sin2B)Fromsinetheoremwehave:asinA=bsinB=csinC=2R⇒a2=4R2sin2Ab2=4R2sin2B,c2=4R2sim2C.ReplaceintoaboveequalitywegetΣa2(sin2B−sin2C)=4R2[sin2A(sin2B−sin2C)+sin2B(sin2C−sin2A)+sin2C(sin2A−sin2B)].(Putx=sin2A,y=sin2B,z=sin2C)=4R2[x(y−z)+y(z−x)+z(x−y)]=4R2(xy−xz+yz−yx+zx−zy)=0

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