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 $3B = 180^{\circ}$,so $B = 60^{\ci

Vedclass · Accepted Answer

$\frac{a+c}{2}$

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 $3B = 180^{\circ}$,so $B = 60^{\circ}$.Using the Law of Cosines,$b^2 = a^2 + c^2 - 2ac \cos B$. Since $B = 60^{\circ}$,$\cos 60^{\circ} = 1/2$,so $b^2 = a^2 + c^2 - ac$.Thus,$\sqrt{a^2 - ac + c^2} = b$.Using the Projection Formula,$b = a \cos C + c \cos A$.Alternatively,using the Mollweide's formula or the ratio of sides,we know $a = 2R \sin A$ and $c = 2R \sin C$.Then $\sqrt{a^2 - ac + c^2} = 2R \sqrt{\sin^2 A - \sin A \sin C + \sin^2 C}$.Using the identity $\cos((A-C)/2) = \sin((A+C)/2) = \sin(180^{\circ} - B/2) = \sin(180^{\circ} - 60^{\circ}/2) = \sin 120^{\circ}$ is not correct here.Actually,$\cos((A-C)/2) = \frac{a+c}{b} \sin(B/2) = \frac{a+c}{b} \sin 30^{\circ} = \frac{a+c}{2b}$.Therefore,$\sqrt{a^2 - ac + c^2} \cdot \cos \left(\frac{A-C}{2}\right) = b \cdot \frac{a+c}{2b} = \frac{a+c}{2}$.

In $\triangle ABC$,if $A, B, C$ are in arithmetic progression,then $\sqrt{a^2-ac+c^2} \cdot \cos \left(\frac{A-C}{2}\right) =$

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