Quadrilateral ABCD is inscribed in circle, $AB=4, BC=5, CD=6, DA=7$, how long is $AC$?

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Quadrilateral ABCD is inscribed in circle, $AB=4, BC=5, CD=6, DA=7$, how long is $AC$?

I think I'm probably supposed to use Ptolemy's to solve this, but I don't know if it's possible. Is there a way to do this problem using Ptolemy's?

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1 Answer

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By the theorem of cosines we get$$AC^2=4^2+5^2-2\times 4\times 5\cos(\beta)$$$$AC^2=7^2+6^2-2\times7\times6\cos(180^{\circ}-\beta)$$ and $$\cos(\pi-x)=-\cos(x)$$

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