In a triangle ABC,if b=7, c=4sqrt{3} and A=fr | vedclass

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(D) Given: $A=\frac{\pi}{6}, b=7, c=4\sqrt{3}$.By the cosine rule,$a^2 = b^2 + c^2 - 2bc \cos A$.$a^2 = 7^2 + (4\sqrt{3})^2 - 2(7)(4\sqrt{3}) \cos(\fr

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$\frac{7\sqrt{3}}{\sqrt{13}}$

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(D) Given: $A=\frac{\pi}{6}, b=7, c=4\sqrt{3}$.By the cosine rule,$a^2 = b^2 + c^2 - 2bc \cos A$.$a^2 = 7^2 + (4\sqrt{3})^2 - 2(7)(4\sqrt{3}) \cos(\frac{\pi}{6})$.$a^2 = 49 + 48 - 56\sqrt{3} 	imes \frac{\sqrt{3}}{2} = 97 - 84 = 13$.So,$a = \sqrt{13}$.By the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.Thus,$\sin B = \frac{b \sin A}{a} = \frac{7 \sin(\pi/6)}{\sqrt{13}} = \frac{7}{2\sqrt{13}}$.And $\sin C = \frac{c \sin A}{a} = \frac{4\sqrt{3} \sin(\pi/6)}{\sqrt{13}} = \frac{2\sqrt{3}}{\sqrt{13}}$.Therefore,$a \sin B \sin C = \sqrt{13} 	imes \frac{7}{2\sqrt{13}} 	imes \frac{2\sqrt{3}}{\sqrt{13}} = \frac{7\sqrt{3}}{\sqrt{13}}$.

In a triangle $ABC$,if $b=7, c=4\sqrt{3}$ and $A=\frac{\pi}{6}$,then $a \sin B \sin C =$

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