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Can you solve this?

 

In quadrilateral $BCED$, sides $\overline{BD}$ and $\overline{CE}$ are extended past $B$ and $C$, respectively, to meet at point $A$. If $BD = 18$, $BC = 18$, $CE = 6$, $AC = 14$ and $AB = 10$, then what is $DE$?

 Apr 2, 2024
 #1
avatar+128845 
+1

Law of Cosines

BC^2  = BA^2  + AC^2  -2 ( BA * AC) cos (DAE)

18^2 = 10^2 + 14^2  - 2 ( 10 * 14)  cos (DAE)

[ 18^2 - 10^2 - 14^2 ] / [ - 2 (10 * 14) ] =  cos (DAE)

-1/10  = cos (DAE)

 

Again

DE^2  =  AD^2 + AE^2 - 2(AD * AE) cos (DAE)

DE^2  = 28^2 + 20^2 - 2(28 * 20) (-1/10)

DE^2  = 1296

DE  =  sqrt [1296]    = 36

 

cool cool cool

 Apr 2, 2024

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