In triangle ABC, with usual notations, if cos | vedclass

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(C) $1$. Given $\cos A = \frac{1}{2}$, so $A = 60^\circ$ or $\frac{\pi}{3}$.$2$. In $	riangle ABC$, $\angle A + \angle B + \angle C = \pi$. Thus, $\a

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$b = \sqrt{3}, c = 2\sqrt{3}$

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(C) $1$. Given $\cos A = \frac{1}{2}$, so $A = 60^\circ$ or $\frac{\pi}{3}$.$2$. In $	riangle ABC$, $\angle A + \angle B + \angle C = \pi$. Thus, $\angle C = \pi - (\frac{\pi}{3} + \frac{\pi}{6}) = \pi - \frac{\pi}{2} = \frac{\pi}{2}$.$3$. Using the Sine Rule, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.$4$. $\frac{3}{\sin(\pi/3)} = \frac{b}{\sin(\pi/6)} = \frac{c}{\sin(\pi/2)}$.$5$. $\frac{3}{\sqrt{3}/2} = \frac{b}{1/2} = \frac{c}{1}$.$6$. $2\sqrt{3} = 2b = c$. Therefore, $b = \sqrt{3}$ and $c = 2\sqrt{3}$.

In $\triangle ABC$, with usual notations, if $\cos A = \frac{1}{2}$, $a = 3$, and $\angle B = \frac{\pi}{6}$, then the values of $b$ and $c$ are:

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