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(B) Let the sides be $a = 7$,$b = 4\sqrt{3}$,and $c = \sqrt{13}$.Since $c$ is the smallest side,the smallest angle is $C$.Using the law of cosines: $\

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$\frac{\pi}{6}$

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(B) Let the sides be $a = 7$,$b = 4\sqrt{3}$,and $c = \sqrt{13}$.Since $c$ is the smallest side,the smallest angle is $C$.Using the law of cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.Substituting the values: $\cos C = \frac{7^2 + (4\sqrt{3})^2 - (\sqrt{13})^2}{2 	imes 7 	imes 4\sqrt{3}}$.$\cos C = \frac{49 + 48 - 13}{56\sqrt{3}} = \frac{84}{56\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.Therefore,$C = \cos^{-1}\left(\frac{\sqrt{3}}{2}ight) = \frac{\pi}{6}$.

With usual notations,if the lengths of the sides of a triangle are $7 \text{ cm}$,$4\sqrt{3} \text{ cm}$,and $\sqrt{13} \text{ cm}$,then the measure of the smallest angle is

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