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

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 $$\cos(\pi-x)=-\cos(x)$$
