# YW bisects <XYZ of triangle XYZ. If YX=4, YZ=5, and XZ=6, find WZ.

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### 2 Answers

In the triangle XYZ, YX=4, YZ=5, and XZ=6. The line YW bisects the angle XYZ.

Use the law of cosines to determine the angle XYZ,

`XZ^2 = XY^2 + YZ^2 - 2*XY*YZ*cos XYZ`

=> `6^2 = 4^2 + 5^2 - 2*4*5*cos XYZ`

=> `cos XYZ = (16 + 25 - 36)/40 = 1/8`

`XYZ = cos^-1(1/8) ~~ 82.82 ` ``degrees

Similarly for angle XZY, `cos XZY = (5^2 + 6^2 - 4^2)/(2*6*5) = 3/4`

`XZY = cos^-1(3/4) ~~ 41.41` degrees

In the triangle WYZ, the angle `WYZ = (1/2)*(XYZ) ~~ 41.41` , WZY = 41.41 and ZY = 5. The angle ZWY = 180 - 41.41 - 41.41 = 97.18 degrees.

Use the law of sines

`(WZ)/(sin WYZ) = (ZY)/(sin ZWY)`

=> `(WZ)/(sin 41.41) = 5/(sin 97.18)`

=> `(WZ) = (5*sin 41.41)/(sin 97.18) = 10/3`

**The length of side WZ = 10/3**

**Sources:**

`Apply ``cos`` law `

`c^2=a^2+b^2-2abcos(C)`

` y^2=x^2+z^2-2xzcos(Y)`

` 2xzcos(Y)=x^2+z^2-y^2 `

`cos(Y)=(x^2+z^2-y^2)/(2xz)`

` =(5^2+4^2-6^2)/(2.5.4) `

`=(41-36)/40 `

`=5/40 `

`=1/8 `

`Y =cos^(-1)(1/8) `

`=82.82 `

`Y/2=41.41`

` and `

`Cos(X)=(y^2+z^2-x^2)/(2yz) `

`=(36+16-25)/(2.6.4)`

` =27/48 =.56 `

`X=cos^(-1)(27/48)`

` =55.77 `

`angle Z=180-55.77-82.82 Z`

`=41.41 `

`We know angle sum law of triangle In Triangle YXW,`

` angle W +angle (Y/2)+ angle Z=180 `

`angle W+41.41+41.41=180 `

`angle W=97.18`

` By sin Law `

`(wz)/sin(Y/2)=(yz)/sin(W) `

`wz= (5. sin(Y/2))/(sin(W) )`

`=(5.sin(41.41))/(sin(97.41))`

` =3.34 `

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