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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+9664 
+1

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

\(\tan^{-1} \dfrac{k}{15} + \tan^{-1} \dfrac{3k}{20} = 60^\circ\)

Taking tan on both sides and using compound angle formula,

\(\dfrac{\frac k{15} + \frac{3k}{20}}{1 - \frac{k}{15} \cdot \frac{3k}{20}} = \sqrt 3\\ \dfrac{\sqrt 3 k}{45} + \dfrac{\sqrt 3 k}{20} = 1- \dfrac{k^2}{100}\\ \dfrac{13\sqrt 3k}{180} = 1 - \dfrac{k^2}{100}\\ k^2 - 100 = -\dfrac{65 \sqrt 3 k}9\\ k^2 + \dfrac{65 \sqrt 3}9k - 100 = 0\\ \left(k + \dfrac{65 \sqrt 3}{18}\right)^2 = 100 + \left(\dfrac{65 \sqrt 3}{18}\right)^2 = \dfrac{15025}{108}\\ k = -\dfrac{65 \sqrt 3}{18} \pm \dfrac{5 \sqrt {601}}{6 \sqrt 3} \text{(reject negative)}\\ k = -\dfrac{65 \sqrt 3}{18} + \dfrac{5 \sqrt{1803}}{18}\)

 

Hence, by Pythagoras' theorem, 

\(AB = \sqrt{(20 - 15)^2 + k^2} = \dfrac{5}{18} \sqrt{2634 - 78 \sqrt{601}}\)

 Apr 8, 2024
 #2
avatar+128794 
+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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