In a Delta ABC,if b = 2,C = 60^circ,and c = s | vedclass

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(C) Using the Law of Sines: $\frac{b}{\sin B} = \frac{c}{\sin C}$.$\sin B = \frac{b \sin C}{c} = \frac{2 \sin 60^\circ}{\sqrt{6}} = \frac{2}{\sqrt{6}}

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$\sqrt{3} + 1$

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(C) Using the Law of Sines: $\frac{b}{\sin B} = \frac{c}{\sin C}$.$\sin B = \frac{b \sin C}{c} = \frac{2 \sin 60^\circ}{\sqrt{6}} = \frac{2}{\sqrt{6}} 	imes \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{\sqrt{6}} = \frac{1}{\sqrt{2}}$.Thus,$B = 45^\circ$ (since $B + C Then $A = 180^\circ - (B + C) = 180^\circ - (45^\circ + 60^\circ) = 75^\circ$.Using the Law of Sines again: $\frac{a}{\sin A} = \frac{c}{\sin C}$.$a = \frac{c \sin A}{\sin C} = \frac{\sqrt{6} \sin 75^\circ}{\sin 60^\circ}$.Since $\sin 75^\circ = \sin(45^\circ + 30^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4}$.$a = \frac{\sqrt{6} 	imes (\frac{\sqrt{6} + \sqrt{2}}{4})}{\frac{\sqrt{3}}{2}} = \frac{6 + \sqrt{12}}{4} 	imes \frac{2}{\sqrt{3}} = \frac{6 + 2\sqrt{3}}{2\sqrt{3}} = \frac{3}{\sqrt{3}} + 1 = \sqrt{3} + 1$.

In a $\Delta ABC$,if $b = 2$,$C = 60^\circ$,and $c = \sqrt{6}$,then $a =$

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