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 \( DE\)?
+128475 See the following image
We can find angle BAC with the Law of Cosines
We have
BC^2 = AB^2 + AC^2 - 2 ( AB * AC) cos(BAC)
9^2 = 5^2 + 8^2 - 2 (5*8)cos(BAC)
[ 9^2 - 5^2 - 8^2]
______________ = cos BAC = 1/10
[ -2 * 5 * 8 ]
So using the Law of Cosines, again we have that
DE = √[ AE^2 + AD^2 - 2 (AE * AD) cos (BAC)]
DE = √[ 10^2 + 16^2 - 2 (10*16)(1/10) ]
DE = √ [ 100 + 256 - 32]
DE = √324 = 18
