In triangle ABC,if (a+c)^2 = b^2 + 3ca,then f | vedclass

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(B) Given $(a+c)^2 = b^2 + 3ca$,we expand the left side: $a^2 + c^2 + 2ac = b^2 + 3ca$. Rearranging gives $a^2 + c^2 - b^2 = ca$. Using the Law of Cos

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$\sqrt{3} \cos \left(\frac{A-C}{2}\right)$

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(B) Given $(a+c)^2 = b^2 + 3ca$,we expand the left side: $a^2 + c^2 + 2ac = b^2 + 3ca$. Rearranging gives $a^2 + c^2 - b^2 = ca$. Using the Law of Cosines,$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{ca}{2ac} = \frac{1}{2}$. Thus,$B = 60^{\circ}$,which implies $\frac{B}{2} = 30^{\circ}$. Using the Sine Rule,$\frac{a}{2R} = \sin A$ and $\frac{c}{2R} = \sin C$,so $\frac{a+c}{2R} = \sin A + \sin C$. Using the sum-to-product formula,$\sin A + \sin C = 2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right)$. Since $A+B+C = 180^{\circ}$,$\frac{A+C}{2} = 90^{\circ} - \frac{B}{2}$. Therefore,$\frac{a+c}{2R} = 2 \sin \left(90^{\circ} - \frac{B}{2}\right) \cos \left(\frac{A-C}{2}\right) = 2 \cos \left(\frac{B}{2}\right) \cos \left(\frac{A-C}{2}\right)$. Substituting $B/2 = 30^{\circ}$,we get $2 \cos 30^{\circ} \cos \left(\frac{A-C}{2}\right) = 2 \left(\frac{\sqrt{3}}{2}\right) \cos \left(\frac{A-C}{2}\right) = \sqrt{3} \cos \left(\frac{A-C}{2}\right)$.

In $\triangle ABC$,if $(a+c)^2 = b^2 + 3ca$,then $\frac{a+c}{2R} =$

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