+743 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 $AB$.
+129881 W A 15 X
x
60 C
3x
Z B 20 Y
Let ACX = Angle A
Let BCY = Angle B
A + B + 60 = 180
A + B = 120
tan ACX = 15/x
tan BCY = 20 / (3x)
tan (A + B) = tan (120)
[ tan A + tan B] / 1 - tanAtanB = -sqrt 3
[ 15/x + 20/(3x) ] / [ 1 - (15/x)(20/(3x)) ] = -sqrt 3
[ 15 (3x) + 20x ] / [ 3x^2 ]
_____________________ = -sqrt (3)
1 - 300 / (3x^2 )
[ 45x + 20x ] / [3x^2]
________________ = -sqrt 3
[3x^2 - 300] / [3x^2]
65x = sqrt (3) [300 - 3x^2]
3sqrt (3)x^2 + 65x - 300sqrt (3) = 0
Using the Quad Formula and taking the positive value of x ≈ 5.5403
3x = 16.6209
AC = sqrt [ 15^2 + (5.5403)^2 ] ≈ sqrt [ 225.69 ]
BC = sqrt [ 20^2 + (16.6209)^2 ] ≈ sqrt [676.25]
Law of Cosines
AB^2 = AC^2 + BC^2 - 2 (AC * BC) cos (60)
AB^2 = 225.69 + 676.25 - 2 sqrt (225.69*676.25 ] * (1/2)
AB^2 = 225.69 + 676.25 - sqrt [ 225.69* 676.25 ]
AB^2 = 511.270
AB = sqrt (511.270) ≈ 22.6
