A Point O is the centre of a circle circumscribed about a triangle ABC- Then OA sin 2 A - OB sin 2 B - OC sin 2 C is equal toA- OA - OB - OC sin 2 AB- 3 OG- where G is the centroid of triangle ABCC- 0D- None of these

The correct option is C 0Let V = nbsp; OAsin 2A+OBsin 2B+OCsin 2C Taking dot product with OA, we get V·OA= OA·OAsin 2A+OB·OAsin 2B +OC·OBsin 2C ....(1) ...

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Byju's Answer

Standard XII

Mathematics

Ratios of Distances between Centroid, Circumcenter, Incenter and Orthocenter of Triangle

A Point O is ...

Question

A Point $O$ is the centre of a circle circumscribed about a triangle $A B C$ . Then $- - \to O A sin 2 A + - - \to O B sin 2 B + - - \to O C sin 2 C$ is equal to

(- - \to O A + - - \to O B + - - \to O C) sin 2 A

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3 - - \to O G

, where

G

is the centroid of triangle

A B C

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\to 0

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None of these

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Solution

The correct option is C $\to 0$
Let
$\to V = - - \to O A sin 2 A + - - \to O B sin 2 B + - - \to O C sin 2 C$
Taking dot product with $- - \to O A$ , we get
$\to V \cdot - - \to O A = - - \to O A \cdot - - \to O A sin 2 A + - - \to O B \cdot - - \to O A sin 2 B + - - \to O C \cdot - - \to O B sin 2 C . . . . (1)$

Now $- - \to O B \cdot - - \to O A = R^{2} cos 2 C$
Similarly,
$- - \to O C \cdot - - \to O A = R^{2} cos 2 B$
and $- - \to O A \cdot - - \to O A = R^{2}$

Putting values in eqn $(1)$

$\to V \cdot - - \to O A = R^{2} sin 2 A + R^{2} cos 2 C sin 2 B + R^{2} cos 2 B sin 2 C$
$\Rightarrow R^{2} sin 2 A + R^{2} sin (2 B + 2 C) . . . (2)$

Now $A + B + C = π$
$\Rightarrow 2 A + 2 B + 2 C = 2 π$
$sin (2 B + 2 C) = sin (2 π - 2 A) = - sin 2 A$
Putting this value in $(2)$
$\to V \cdot - - \to O A = 0 \to \to V = 0$ or $\to V ⊥ - - \to O A$
Similarly,
$\to V \cdot - - \to O B = 0 \to \to V = 0$ or $\to V ⊥ - - \to O B$
$\to V \cdot - - \to O C = 0 \to \to V = 0$ or $\to V ⊥ - - \to O C$

$\to V = λ_{1} - - \to O A + λ_{2} - - \to O B + λ_{3} - - \to O C$
$\to V$ is coplanar and
$\to V$ is $⊥$ to $- - \to O B, - - \to O B, - - \to O C$
Which is not possible
So $\to V = 0$

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A Point O is the centre of a circle circumscribed about a triangle ABC. Then −−→OAsin2A+−−→OBsin2B+−−→OCsin2C is equal to

A Point $O$ is the centre of a circle circumscribed about a triangle $A B C$ . Then $- - \to O A sin 2 A + - - \to O B sin 2 B + - - \to O C sin 2 C$ is equal to