In any triangle ABC,b^2 sin 2C + c^2 sin 2B i | vedclass

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(D) We are given the expression $b^2 \sin 2C + c^2 \sin 2B$. Using the double angle formula $\sin 2	heta = 2 \sin 	heta \cos 	heta$,we get: $= b^2

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$4 \Delta$

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(D) We are given the expression $b^2 \sin 2C + c^2 \sin 2B$. Using the double angle formula $\sin 2	heta = 2 \sin 	heta \cos 	heta$,we get: $= b^2 (2 \sin C \cos C) + c^2 (2 \sin B \cos B)$ Using the Sine Rule,$\sin C = \frac{c}{2R}$ and $\sin B = \frac{b}{2R}$: $= 2b^2 \left(\frac{c}{2R}ight) \cos C + 2c^2 \left(\frac{b}{2R}ight) \cos B$ $= \frac{b^2 c \cos C}{R} + \frac{c^2 b \cos B}{R} = \frac{bc}{R} (b \cos C + c \cos B)$ By the projection formula,$b \cos C + c \cos B = a$: $= \frac{bc}{R} (a) = \frac{abc}{R}$ Since $R = \frac{abc}{4\Delta}$,it follows that $\frac{abc}{R} = 4\Delta$.

In any $\triangle ABC$,$b^2 \sin 2C + c^2 \sin 2B$ is equal to

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