+921 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 $AB$.
+9519 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,
\(AB = \sqrt{(20 - 15)^2 + k^2} = \dfrac{5}{18} \sqrt{2634 - 78 \sqrt{601}}\)
+128485 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
