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 $AB$.
Let CX=k. Then, since AX is parallel to BY, we have
tan−1k15+tan−13k20=60∘
Taking tan on both sides and using compound angle formula,
k15+3k201−k15⋅3k20=√3√3k45+√3k20=1−k210013√3k180=1−k2100k2−100=−65√3k9k2+65√39k−100=0(k+65√318)2=100+(65√318)2=15025108k=−65√318±5√6016√3(reject negative)k=−65√318+5√180318
Hence, by Pythagoras' theorem,
AB=√(20−15)2+k2=518√2634−78√601
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