Question

# If point O is the centre of a circle circumscribed about a triangle ABC. Then Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯OAsin2A+Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯OBsin2B+Â¯Â¯Â¯Â¯Â¯Â¯Â¯Â¯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 ¯¯¯0In 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)=√32So after solving we get ¯¯¯¯¯¯¯¯OA + ¯¯¯¯¯¯¯¯OB + ¯¯¯¯¯¯¯¯OC =0.Hence the answer comes out to be 0.

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