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

Question

A point O is the centre of a circle circumscribed about a triangle ABC. Then OA sin 2A + OB sin 2B + OC sin 2C is equal to (A) (OA+OB+OC) sin 2A (B) 3 OG , where G is the centroid of triangle ABC (C)  vec 0 (D) none of these

Tardigrade Admin · Accepted Answer

( vecOA sin 2A + vec OB sin 2B + vecOC sin 2C/sin 2A + sin 2B + sin 2C) ⇒ ( vecOA sin 2A + vec OB sin 2B + vecOC sin 2C/sin 2A + sin 2B + sin 2C)= vecO or OA sin 2 A + OB sin 2B + OC sin 2C = vec0

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

$(O A + OB + OC)$ $s in 2 A$

$3 OG$ , where $G$ is the centroid of triangle $A BC$

$0$

none of these

$\frac{O A s in 2 A + OB s in 2 B + OC s in 2 C}{s in 2 A + s in 2 B + s in 2 C}$
$\Rightarrow \frac{O A s in 2 A + OB s in 2 B + OC s in 2 C}{s in 2 A + s in 2 B + s in 2 C} = O$
or $O A s in 2 A + OB s in 2 B + OC s in 2 C = 0$

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

(OA+OB+OC) sin2A

3OG , where G is the centroid of triangle ABC

0

none of these

sin2A+sin2B+sin2COAsin2A+OBsin2B+OCsin2C​ ⇒sin2A+sin2B+sin2COAsin2A+OBsin2B+OCsin2C​=O or OAsin2A+OBsin2B+OCsin2C=0

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

$(O A + OB + OC)$ $s in 2 A$

$3 OG$ , where $G$ is the centroid of triangle $A BC$

$0$

$\frac{O A s in 2 A + OB s in 2 B + OC s in 2 C}{s in 2 A + s in 2 B + s in 2 C}$
$\Rightarrow \frac{O A s in 2 A + OB s in 2 B + OC s in 2 C}{s in 2 A + s in 2 B + s in 2 C} = O$
or $O A s in 2 A + OB s in 2 B + OC s in 2 C = 0$