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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 $DE$?

 Apr 3, 2020
 #1
avatar+111321 
+3

 

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

 

 

 

cool cool cool

 Apr 3, 2020
 #2
avatar+55 
+1

Thank you so much!!!

 Apr 3, 2020

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