The angles A, B, C of a triangle ABC are in A | vedclass

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(C) Given that the angles $A, B, C$ of $	riangle ABC$ are in $AP$,we have $A+C = 2B$.Since $A+B+C = 180^{\circ}$,we get $3B = 180^{\circ}$,which impl

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$\sqrt{43}$

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(C) Given that the angles $A, B, C$ of $	riangle ABC$ are in $AP$,we have $A+C = 2B$.Since $A+B+C = 180^{\circ}$,we get $3B = 180^{\circ}$,which implies $B = 60^{\circ}$.Using the cosine rule in $	riangle ABC$:$\cos B = \frac{AB^2 + BC^2 - AC^2}{2 	imes AB 	imes BC}$Substituting the given values $AB=6, BC=7, B=60^{\circ}$ and letting $AC=x$:$\cos 60^{\circ} = \frac{6^2 + 7^2 - x^2}{2 	imes 6 	imes 7}$$\frac{1}{2} = \frac{36 + 49 - x^2}{84}$$42 = 85 - x^2$$x^2 = 85 - 42 = 43$$x = \sqrt{43}$Therefore,$AC = \sqrt{43}$.

The angles $A, B, C$ of a triangle $ABC$ are in $AP$. If $AB=6$ and $BC=7$,then $AC=$

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