+111 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 \(AC = 28\) and \(AB = 24\), then what is \(DE\)?
Any help would be appriciated. Thank you so much!
+23252 Notice that you have two triangles, triangle(ABC) and triangle(ADE).
You can use the Law of Cosines on each of these triangles.
First: triangleABC): BC2 = AB2 + AC2 - 2·AB·AC·cos( angle(A) )
92 = 242 + 282 - 2·24·28·cos( angle(A) )
Solve this for angle(A).
then: triangle(ADE): DE2 = AD2 + AE2 - 2·AD·AE·cos( angle(A) )
DE2 = 352 + 302 - 2·35·30·cos( angle(A) ) (use the value of angle(A) from above)
Solve this for DE.
