Question

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

• A. $\left(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\right)$ $sin\,2A$
• B. $3 \overrightarrow{OG}$ , where $G$ is the centroid of triangle $ABC$
• C. $\vec{ 0}$
• D. none of these

Answer 1

Answer: (C)

$\frac{\vec{OA} \, sin \,2A +\vec{ OB} \, sin \,2B + \vec{OC } \, sin \,2C}{sin \,2A + sin \,2B + sin \, 2C}$
$\Rightarrow \frac{\vec{OA} \, sin \, 2A +\vec{ OB} \, sin \,2B + \vec{OC } \,sin \,2C}{sin \,2A + sin \,2B + sin \,2C}=\vec{O}$
or $\overrightarrow{ OA} \,sin \,2 A + \overrightarrow{ OB} \, sin \, 2B + \overrightarrow{OC} \,sin \,2C = \vec{0}$