In triangle ABC,if A + C = 2B,then frac{a + c | vedclass

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(A) Given $A + C = 2B$. Since $A + B + C = 180^{\circ}$,we have $3B = 180^{\circ}$,so $B = 60^{\circ}$.Using the Law of Cosines,$\cos B = \frac{a^2 +

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$2\cos \frac{A - C}{2}$

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(A) Given $A + C = 2B$. Since $A + B + C = 180^{\circ}$,we have $3B = 180^{\circ}$,so $B = 60^{\circ}$.Using the Law of Cosines,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.Since $\cos 60^{\circ} = \frac{1}{2}$,we have $\frac{1}{2} = \frac{a^2 + c^2 - b^2}{2ac}$,which implies $ac = a^2 + c^2 - b^2$,or $b^2 = a^2 - ac + c^2$.Thus,$\sqrt{a^2 - ac + c^2} = b$.The expression becomes $\frac{a + c}{b}$.By the Sine Rule,$\frac{a}{\sin A} = \frac{c}{\sin C} = \frac{b}{\sin B} = 2R$,so $a = 2R \sin A$,$c = 2R \sin C$,and $b = 2R \sin B$.Substituting these,we get $\frac{2R(\sin A + \sin C)}{2R \sin B} = \frac{\sin A + \sin C}{\sin B}$.Using the sum-to-product formula,$\sin A + \sin C = 2 \sin \frac{A + C}{2} \cos \frac{A - C}{2}$.Since $A + C = 120^{\circ}$,$\frac{A + C}{2} = 60^{\circ}$,so $\sin \frac{A + C}{2} = \sin 60^{\circ} = \frac{\sqrt{3}}{2}$.Also,$\sin B = \sin 60^{\circ} = \frac{\sqrt{3}}{2}$.Thus,the expression is $\frac{2 \sin 60^{\circ} \cos \frac{A - C}{2}}{\sin 60^{\circ}} = 2 \cos \frac{A - C}{2}$.

In triangle $ABC$,if $A + C = 2B$,then $\frac{a + c}{\sqrt{a^2 - ac + c^2}}$ is equal to

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