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(B) Given that the angles $A, B, C$ of $	riangle ABC$ are in $A$.$P$.,we have $2B = A + C$. Since $A + B + C = 180^{\circ}$,we get $3B = 180^{\circ}$

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$\sqrt{3}$

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(B) Given that the angles $A, B, C$ of $	riangle ABC$ are in $A$.$P$.,we have $2B = A + C$. Since $A + B + C = 180^{\circ}$,we get $3B = 180^{\circ}$,so $B = 60^{\circ}$ and $A + C = 120^{\circ}$. Using the Sine Rule,$\frac{a}{\sin A} = \frac{c}{\sin C} = k$,so $a = k \sin A$ and $c = k \sin C$. Substituting these into the expression: $\frac{a}{c} \sin 2C + \frac{c}{a} \sin 2A = \frac{k \sin A}{k \sin C} (2 \sin C \cos C) + \frac{k \sin C}{k \sin A} (2 \sin A \cos A)$ $= 2 \sin A \cos C + 2 \sin C \cos A$ $= 2(\sin A \cos C + \cos A \sin C)$ $= 2 \sin(A + C)$ $= 2 \sin(120^{\circ}) = 2 	imes \frac{\sqrt{3}}{2} = \sqrt{3}$.

If in $\triangle ABC$,with usual notations,the angles are in $A$.$P$.,then $\frac{a}{c} \sin 2C + \frac{c}{a} \sin 2A =$

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