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(B) Let the sides be $a = 7$,$b = 4\sqrt{3}$,and $c = \sqrt{13}$.Since $\sqrt{13} \approx 3.6$ and $4\sqrt{3} \approx 6.9$,the smallest side is $c = \

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$30$

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(B) Let the sides be $a = 7$,$b = 4\sqrt{3}$,and $c = \sqrt{13}$.Since $\sqrt{13} \approx 3.6$ and $4\sqrt{3} \approx 6.9$,the smallest side is $c = \sqrt{13}$.The smallest angle is opposite the smallest side,so we find angle $C$ using the Law of Cosines:$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$\cos C = \frac{7^2 + (4\sqrt{3})^2 - (\sqrt{13})^2}{2 	imes 7 	imes 4\sqrt{3}}$$\cos C = \frac{49 + 48 - 13}{56\sqrt{3}}$$\cos C = \frac{84}{56\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$Since $\cos C = \frac{\sqrt{3}}{2}$,we have $C = 30^o$.

If the lengths of the sides of a triangle are $7 \, cm$,$4\sqrt{3} \, cm$,and $\sqrt{13} \, cm$,then the smallest angle is .....$^o$.

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