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

(¯ ¯¯¯¯¯¯ ¯ O A + ¯ ¯¯¯¯¯¯ ¯ O B + ¯ ¯¯¯¯¯¯ ¯ O C) sin 2 A

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3 ¯ ¯¯¯¯¯¯¯ ¯ O G

, where G is the centroid of triangle ABC

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\to 0

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(¯ ¯¯¯¯¯¯ ¯ O A + ¯ ¯¯¯¯¯¯ ¯ O B + ¯ ¯¯¯¯¯¯ ¯ O C) sin 2 B

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Solution

The correct option is C $\to 0$
We know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point
Hence, if $A, B$ and $C$ denotes $∠ A, ∠ B$ and $∠ C$ respectively.
$∠ A O B = 2 C$ , $∠ B O C = 2 A$ and $∠ C O A = 2 B$
Now $∣ ∣ \to O A ∣ ∣ = ∣ ∣ \to O B ∣ ∣ = ∣ ∣ \to O C ∣ ∣ = r$ units
Now let:
$\to v = \to O A sin (2 A) + \to O B sin (2 B) + \to O C sin (2 C)$ ... $(1)$
Taking dot product with $\to O A$ on both sides:
$\to v . \to O A = \to O A . \to O A sin (2 A) + \to O B . \to O A sin (2 B) + \to O C . \to O A sin (2 C)$
As $\to P . \to Q = ∣ ∣ \to P ∣ ∣ . ∣ ∣ \to Q ∣ ∣ cos θ$
$\to O B . \to O A = ∣ ∣ \to O B ∣ ∣ . ∣ ∣ \to O A ∣ ∣ cos 2 C$
$\to O C . \to O A = ∣ ∣ \to O C ∣ ∣ . ∣ ∣ \to O A ∣ ∣ cos 2 B$
Hence, $\to v . \to O A = r^{2} sin 2 A + r^{2} cos 2 C sin 2 B + r^{2} cos 2 B sin 2 C$
$= r^{2} [sin 2 A + (cos 2 C sin 2 B + cos 2 B sin 2 C)]$
$= r^{2} [sin 2 A + sin (2 B + 2 C)]$
$= r^{2} [sin 2 A + sin ((2 π - 2 A))]$ since sum of all angles of triangle $= π$
$= r^{2} [sin 2 A - sin 2 A]$ since $sin (2 π - θ) = - sin θ$
$= r^{2} \times 0 = 0$
$∴ \to v . \to O A = 0$
Taking dot product in eqn $(1)$ with $\to O B$ on both sides:
$\to v . \to O B = \to O A . \to O B sin (2 A) + \to O B . \to O B sin (2 B) + \to O C . \to O B sin (2 C)$
As $\to P . \to Q = ∣ ∣ \to P ∣ ∣ . ∣ ∣ \to Q ∣ ∣ cos θ$
$\Rightarrow \to O A . \to O B = ∣ ∣ \to O A ∣ ∣ . ∣ ∣ \to O B ∣ ∣ cos 2 C$
$\Rightarrow \to O C . \to O B = ∣ ∣ \to O C ∣ ∣ . ∣ ∣ \to O B ∣ ∣ cos 2 A$
Hence, $\to v . \to O A = r^{2} cos 2 C sin 2 A + r^{2} sin 2 B + r^{2} cos 2 A sin 2 C$
$= r^{2} [sin 2 A + (cos 2 C sin 2 B + cos 2 A sin 2 C)]$
$= r^{2} [sin 2 B + sin (2 A + 2 C)]$
$= r^{2} [sin 2 B + sin ((2 π - 2 B))]$ since sum of all angles of triangle $= π$
$= r^{2} [sin 2 B - sin 2 B]$ since $sin (2 π - θ) = - sin θ$
$= r^{2} \times 0 = 0$
$∴ \to v . \to O B = 0$
Taking dot product in eqn $(1)$ with $\to O C$ on both sides:
$\to v . \to O C = \to O A . \to O C sin (2 A) + \to O B . \to O C sin (2 B) + \to O C . \to O C sin (2 C)$
As $\to P . \to Q = ∣ ∣ \to P ∣ ∣ . ∣ ∣ \to Q ∣ ∣ cos θ$
$\Rightarrow \to O A . \to O C = ∣ ∣ \to O A ∣ ∣ . ∣ ∣ \to O C ∣ ∣ cos 2 B$
$\Rightarrow \to O B . \to O C = ∣ ∣ \to O B ∣ ∣ . ∣ ∣ \to O C ∣ ∣ cos 2 A$
Hence, $\to v . \to O C = r^{2} cos 2 B sin 2 A + r^{2} cos 2 A sin 2 B + r^{2} sin 2 C$
$= r^{2} [sin 2 C + (cos 2 B sin 2 A + cos 2 A sin 2 B)]$
$= r^{2} [sin 2 C + sin (2 A + 2 B)]$
$= r^{2} [sin 2 C + sin ((2 π - 2 C))]$ since sum of all angles of triangle $= π$
$= r^{2} [sin 2 C - sin 2 C]$ since $sin (2 π - θ) = - sin θ$
$= r^{2} \times 0 = 0$
$∴ \to v . \to O C = 0$
Hence, $\to v . \to O A = \to v . \to O B = \to v . \to O C = 0$
If $\to v \neq 0,$ then $\to v$ must be perpendicular to $\to O A, \to O B$ and $\to O C$ which is impossible.
Hence $\to v = 0$ ,
$\Rightarrow \to O A sin (2 A) + \to O B sin (2 B) + \to O C sin (2 C) = \to 0$

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