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Question

A Point O is the centre of a circle circumscribed about a triangle ABC. Then −−→OAsin2A+−−→OBsin2B+−−→OCsin2C is equal to

A
(OA+OB+OC)sin2A
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B
3OG, where G is the centroid of triangle ABC
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C
0
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D
None of these
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Solution

The correct option is C →0Let →V=−−→OAsin2A+−−→OBsin2B+−−→OCsin2C Taking dot product with −−→OA, we get →V⋅−−→OA=−−→OA⋅−−→OAsin2A+−−→OB⋅−−→OAsin2B+−−→OC⋅−−→OBsin2C ....(1) Now −−→OB⋅−−→OA=R2cos2C Similarly, −−→OC⋅−−→OA=R2cos2B and −−→OA⋅−−→OA=R2 Putting values in eqn (1) →V⋅−−→OA=R2sin2A+R2cos2Csin2B+R2cos2Bsin2C ⇒R2sin2A+R2sin(2B+2C) ...(2) Now A+B+C=π ⇒2A+2B+2C=2π sin(2B+2C)=sin(2π−2A)=−sin2A Putting this value in (2) →V⋅−−→OA=0→→V=0 or →V⊥−−→OA Similarly, →V⋅−−→OB=0→→V=0 or →V⊥−−→OB →V⋅−−→OC=0→→V=0 or →V⊥−−→OC →V=λ1−−→OA+λ2−−→OB+λ3−−→OC →V is coplanar and →V is ⊥ to −−→OB,−−→OB,−−→OC Which is not possible So →V=0

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