If ABCD is a cyclic quadrilateral with AB=6, | vedclass

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(A) In $	riangle ABC$,by the law of cosines:$AC^2 = AB^2 + BC^2 - 2(AB)(BC) \cos 	heta$$AC^2 = 6^2 + 4^2 - 2(6)(4) \cos 	heta = 36 + 16 - 48 \cos \

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$\frac{3}{13}$

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(A) In $	riangle ABC$,by the law of cosines:$AC^2 = AB^2 + BC^2 - 2(AB)(BC) \cos 	heta$$AC^2 = 6^2 + 4^2 - 2(6)(4) \cos 	heta = 36 + 16 - 48 \cos 	heta = 52 - 48 \cos 	heta \quad (i)$In a cyclic quadrilateral,opposite angles are supplementary. Thus,$\angle ADC = 180^{\circ} - 	heta$.In $	riangle ADC$,by the law of cosines:$AC^2 = AD^2 + CD^2 - 2(AD)(CD) \cos(180^{\circ} - 	heta)$Since $\cos(180^{\circ} - 	heta) = -\cos 	heta$,we have:$AC^2 = 3^2 + 5^2 - 2(3)(5)(-\cos 	heta) = 9 + 25 + 30 \cos 	heta = 34 + 30 \cos 	heta \quad (ii)$Equating $(i)$ and $(ii)$:$52 - 48 \cos 	heta = 34 + 30 \cos 	heta$$52 - 34 = 30 \cos 	heta + 48 \cos 	heta$$18 = 78 \cos 	heta$$\cos 	heta = \frac{18}{78} = \frac{3}{13}$

If $ABCD$ is a cyclic quadrilateral with $AB=6, BC=4, CD=5, DA=3$ and $\angle ABC=\theta$,then $\cos \theta=$

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