In Delta ABC,2R^2 sin A sin B sin C = | vedclass

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(C) We know that by the Sine Rule,$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = \frac{1}{2R}$.Therefore,$\sin A = \frac{a}{2R}$,$\sin B =

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$\Delta$

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(C) We know that by the Sine Rule,$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = \frac{1}{2R}$.Therefore,$\sin A = \frac{a}{2R}$,$\sin B = \frac{b}{2R}$,and $\sin C = \frac{c}{2R}$.Substituting these into the expression:$2R^2 \sin A \sin B \sin C = 2R^2 \left( \frac{a}{2R} \right) \left( \frac{b}{2R} \right) \left( \frac{c}{2R} \right)$$= 2R^2 \left( \frac{abc}{8R^3} \right)$$= \frac{abc}{4R}$.Since the area of a triangle $\Delta = \frac{abc}{4R}$,the expression equals $\Delta$.

In $\Delta ABC$,$2R^2 \sin A \sin B \sin C = $

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