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Question

In a triangle ABC with usual notation, which of the following is (are) CORRECT? 
  1. If the angles A,B,C are in A.P., then b2=c2+a2ca
  2. sin(BC)sin(B+C)=b2c2a2
  3. b2c2cosB+cosC=0
  4. 1+cos(AB)cosC1+cos(AC)cosB=a2+b2a2+c2


Solution

The correct options are
A If the angles A,B,C are in A.P., then b2=c2+a2ca
B sin(BC)sin(B+C)=b2c2a2
C b2c2cosB+cosC=0
D 1+cos(AB)cosC1+cos(AC)cosB=a2+b2a2+c2
A,B,C are in A.P.
B=60
Now, by cosine rule,
b2=c2+a22cacosB
=c2+a22ca×12
=c2+a22ca

sin(BC)sin(B+C)
=sin(BC)sin(B+C)sin2(B+C)
=sin2Bsin2Csin2A
=b2c2a2     [By sine rule]

b2c2cosB+cosC
=4R2(sin2Bsin2C)cosB+cosC    [By sine rule]
=4R2(cos2Ccos2B)cosB+cosC
=4R2(cosCcosB)
b2c2cosB+cosC
=4R2(cosCcosB+cosAcosC+cosBcosA)
=0

1+cos(AB)cosC1+cos(AC)cosB
=1cos(AB)cos(A+B)1cos(AC)cos(A+C)
=1(cos2Acos2B)1(cos2Acos2C)
=sin2A+sin2Bsin2A+sin2C
=a2+b2a2+c2

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