In triangle ABC,if A, B, C are in arithmetic | vedclass

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(B) Given that $A, B, C$ are in arithmetic progression,we have $2B = A + C$. Since $A + B + C = 180^{\circ}$,we get $3B = 180^{\circ}$,which implies $

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(B) Given that $A, B, C$ are in arithmetic progression,we have $2B = A + C$. Since $A + B + C = 180^{\circ}$,we get $3B = 180^{\circ}$,which implies $B = 60^{\circ}$. Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$. Thus,$a = 2R \sin A$,$c = 2R \sin C$. The expression becomes: $\frac{c}{a} \sin 2A + \frac{a}{c} \sin 2C = \frac{\sin C}{\sin A} (2 \sin A \cos A) + \frac{\sin A}{\sin C} (2 \sin C \cos C)$ $= 2 \sin C \cos A + 2 \sin A \cos C$ $= 2 \sin(A + C)$. Since $A + C = 180^{\circ} - B = 180^{\circ} - 60^{\circ} = 120^{\circ}$,$= 2 \sin(120^{\circ}) = 2 	imes \frac{\sqrt{3}}{2} = \sqrt{3}$.

In $\triangle ABC$,if $A, B, C$ are in arithmetic progression,then $\frac{c}{a} \sin 2A + \frac{a}{c} \sin 2C =$

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