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(C) We know that the area of triangle $ABC$ is given by $\Delta = \frac{1}{2}ab \sin C = \frac{1}{2}bc \sin A = \frac{1}{2}ca \sin B$.Using the sine r

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(C) We know that the area of triangle $ABC$ is given by $\Delta = \frac{1}{2}ab \sin C = \frac{1}{2}bc \sin A = \frac{1}{2}ca \sin B$.Using the sine rule, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$, we have $a = 2R \sin A$ and $b = 2R \sin B$.Now, consider the expression $E = a^2 \sin 2B + b^2 \sin 2A$.$E = a^2 (2 \sin B \cos B) + b^2 (2 \sin A \cos A)$.Substituting $a = 2R \sin A$ and $b = 2R \sin B$:$E = (2R \sin A)^2 (2 \sin B \cos B) + (2R \sin B)^2 (2 \sin A \cos A)$.$E = 8R^2 \sin A \sin B (\sin A \cos B + \sin B \cos A)$.$E = 8R^2 \sin A \sin B \sin(A + B)$.Since $A + B + C = 180^{\circ}$, $\sin(A + B) = \sin C$.$E = 8R^2 \sin A \sin B \sin C$.We know $\Delta = \frac{abc}{4R} = \frac{(2R \sin A)(2R \sin B)(2R \sin C)}{4R} = 2R^2 \sin A \sin B \sin C$.Therefore, $E = 4(2R^2 \sin A \sin B \sin C) = 4\Delta$.

If the area of a triangle $ABC$ is $\Delta$, then ${a^2}\sin 2B + {b^2}\sin 2A$ is equal to (in $\Delta$)

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