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Question

If point O is the centre of a circle circumscribed about a triangle ABC. Then ¯¯¯¯¯¯¯¯OAsin2A+¯¯¯¯¯¯¯¯OBsin2B+¯¯¯¯¯¯¯¯OCsin2C=

A
(¯¯¯¯¯¯¯¯OA+¯¯¯¯¯¯¯¯OB+¯¯¯¯¯¯¯¯OC)sin2A
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B
(¯¯¯¯¯¯¯¯OA+¯¯¯¯¯¯¯¯OB+¯¯¯¯¯¯¯¯OC)cos2A
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C
¯¯¯0
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D
(¯¯¯¯¯¯¯¯OA+¯¯¯¯¯¯¯¯OB+¯¯¯¯¯¯¯¯OC)tan2A
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Solution

The correct option is C ¯¯¯0
In order to make the problem easier and take less time to solve it, we take a simpler case and take the triangle to be an equilateral one.
Thus OA=OB=OC and sin2A=sin2B=sin2C=sin(1200)=32
So after solving we get ¯¯¯¯¯¯¯¯OA + ¯¯¯¯¯¯¯¯OB + ¯¯¯¯¯¯¯¯OC =0.
Hence the answer comes out to be 0.

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