+111 In quadrilateral BCED we have BD=11, BC=9, and CE=2. Sides ¯BD and ¯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!
+23254 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.
