In triangle ABC,if a, b, c are in arithmetic | vedclass

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(C) Given $a, b, c$ are in $AP$,so $2b = a + c$. By the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$. Thus,$2 \sin B = \sin

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$5:4$

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(C) Given $a, b, c$ are in $AP$,so $2b = a + c$. By the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$. Thus,$2 \sin B = \sin A + \sin C$. Given $A = 2C$,so $2 \sin B = \sin 2C + \sin C$. Since $A + B + C = 180^{\circ}$,we have $B = 180^{\circ} - (A + C) = 180^{\circ} - 3C$. So,$2 \sin(180^{\circ} - 3C) = \sin 2C + \sin C$. $2 \sin 3C = 2 \sin C \cos C + \sin C$. $2(3 \sin C - 4 \sin^3 C) = \sin C(2 \cos C + 1)$. Since $\sin C 
eq 0$,we have $2(3 - 4 \sin^2 C) = 2 \cos C + 1$. $6 - 8(1 - \cos^2 C) = 2 \cos C + 1$. $6 - 8 + 8 \cos^2 C = 2 \cos C + 1$. $8 \cos^2 C - 2 \cos C - 3 = 0$. $(4 \cos C + 3)(2 \cos C - 1) = 0$. Since $A = 2C$,$C  0$. Thus,$\cos C = \frac{1}{2}$,which means $C = 60^{\circ}$. Then $A = 120^{\circ}$,which is impossible as $A+C Wait,re-evaluating: $8 \cos^2 C - 2 \cos C - 3 = 0$ gives $\cos C = \frac{2 \pm \sqrt{4 + 96}}{16} = \frac{2 \pm 10}{16}$. Taking $\cos C = \frac{12}{16} = \frac{3}{4}$. Then $\sin C = \sqrt{1 - (\frac{3}{4})^2} = \frac{\sqrt{7}}{4}$. $B = 180^{\circ} - 3C$,so $\sin B = \sin 3C = 3 \sin C - 4 \sin^3 C = \sin C(3 - 4 \sin^2 C) = \sin C(3 - 4(1 - \cos^2 C)) = \sin C(4 \cos^2 C - 1)$. $\sin B = \frac{\sqrt{7}}{4} (4(\frac{9}{16}) - 1) = \frac{\sqrt{7}}{4} (\frac{9}{4} - 1) = \frac{\sqrt{7}}{4} 	imes \frac{5}{4} = \frac{5\sqrt{7}}{16}$. Finally,$b:c = \sin B : \sin C = \frac{5\sqrt{7}}{16} : \frac{\sqrt{7}}{4} = 5:4$.

In $\triangle ABC$,if $a, b, c$ are in arithmetic progression and $A=2C$,then $b:c=$

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