Question

If point $O$ is the centre of a circle circumscribed about a triangle $ABC$. Then $\overline{OA}\sin 2A+\overline{OB}\sin 2B+\overline{OC}\sin 2C=$

• A. $(\overline{OA}+\overline{OB}+\overline{OC})\sin 2A$
• B. $(\overline{OA}+\overline{OB}+\overline{OC})\cos 2A$
• C. $\overline{0}$
• D. $(\overline{OA}+\overline{OB}+\overline{OC})\tan 2A$

Answer 1

The correct option is C $\overline{0}$

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 $sin2A=sin2B=sin2C=sin\left({120}^0\right)=\frac{\sqrt{3}}{2}$
So after solving we get $\overline{OA}$ + $\overline{OB}$ + $\overline{OC}$ $=0.$
Hence the answer comes out to be $0.$