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In rectangle $WXYZ$, $A$ is on side $\overline{WX}$, $B$ is on side $\overline{YZ}$, and $C$ is on side $\overline{XY}$. If $AX = 15$, $BY = 20$, $\angle ACB= 60^\circ$, and $CY = 3 \cdot CX$, then find $AB$.

 Apr 8, 2024
 #1
avatar+9675 
+1

Let CX=k. Then, since AX is parallel to BY, we have

tan1k15+tan13k20=60

Taking tan on both sides and using compound angle formula,

k15+3k201k153k20=33k45+3k20=1k2100133k180=1k2100k2100=653k9k2+6539k100=0(k+65318)2=100+(65318)2=15025108k=65318±560163(reject negative)k=65318+5180318

 

Hence, by Pythagoras' theorem, 

AB=(2015)2+k2=518263478601

 Apr 8, 2024
 #2
avatar+130466 
+1

W               A   15      X

                                 m

                                 C

 

                                3m                            

 

Z         B      20         Y

 

 

tan  CAX =  m / 15   

tan  CBY  = 3m/ 20

 

Since angle ACB   = 60   then   angles CAX + CBY = 120

 

So

 

tan  ( CAX + CBY)  =   tan (120)

 

Identity  

 

tan (A + B)  =  [ tan A + tan B ] / [ 1 - tan A * tan B ] 

 

[ m/15 + (3m) / 20 ] / [ 1 - (m/15) [ (3m) / 20] ] =       -sqrt 3

 

Solving this  for positive m produces  m ≈  18.05  .....3m ≈  54.15

 

AC =  sqrt [ 15^2 + 18.05^2 ] ≈  23.47

BC =  sqrt [ 20^2 + 54.15^2 ] ≈ 57.73

 

Law of Cosines

 

AB  =  sqrt [ 23.47^2 + 57.73^2  - 2 ( 23.47 * 57.73)(1/2) ]  ≈  50.29

 

cool cool cool

 Apr 9, 2024

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