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# Rectangle

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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 $AB$.

Mar 26, 2024

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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

Mar 26, 2024