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