+73 In quadrilateral $BCED$, we have $BD = 11$, $BC = 9$, and $CE=2$. Sides $\overline{BD}$ and $\overline{CE}$ are extended past $B$ and $C$, respectively, to meet at point $A$. If $AC = 8$ and $AB = 5$, then what is $DE$?
+128474 
Using the Law of Cosines we have that
BC^2 = AB^2 + AC^2 - 2(AB* AC) cos ( BAC)
9^2 = 5^2 + 8^2 - 2(5 * 8) cos (BAC)
81 = 25 + 64 - 80 cos (BAC)
cos ( BAC) = [ 81 - 25 - 64 ] / -80 = -8 / -80 = 1/10
And applying it once more we have that
DE^2 = AE^2 + AD^2 - 2(AE * AD) cos (BAC)
DE^2 = 10^2 + 16^2 - 2 ( 10 * 16) (1/10)
DE^2 = 100 + 256 - 32
DE^2 = 356 - 32
DE^2 = 324
DE = 18
