Compute $QY$ from $XY$ and $XQ$. Also, if you let $QZ=z$, then $WQ$ can be written using $WZ=120$.

Your equation for $QZ$ may produce two solutions. Use “$Q$ is closer to $Z$ than to $W$” to decide which solution fits the diagram.

Compute $QY$ as $118-64$, and let $z$ represent $QZ$ so that $WQ=120-z$.

Enter $y=z(120-z)$ and enter $y=54\cdot 64$ as a second equation. The $x$-coordinates of the intersection points are the possible values of $z=QZ$.

Use the condition “$Q$ is closer to $Z$ than to $W$” to pick the smaller value of $z$ (so $QZ < WQ$).

Compute the ratio $\frac{WQ}{QY}$ and then compute $YZ=WX\cdot \frac{QY}{WQ}$. Choose the answer option that equals that result.

Because $Q$ is on $\overline{XY}$,

$QY = XY - XQ = 118 - 64 = 54$.

Triangles $\triangle WQX$ and $\triangle YQZ$ are similar:

• $\angle QWX = \angle QYZ$ (given)
• $\angle WQX = \angle YQZ$ (vertical angles)

So corresponding side ratios are equal:

Let $QZ = z$. Since $WZ = 120$ and $Q$ lies on $\overline{WZ}$,

Substitute $QY=54$ and $XQ=64$:

Cross-multiply:

Rearrange:

So $z=48$ or $z=72$.

If $z=48$, then $WQ = 120-48=72$, so $QZ < WQ$.

If $z=72$, then $WQ = 120-72=48$, so $QZ > WQ$.

Since point $Q$ is closer to $Z$ than to $W$, we need $QZ < WQ$, so $QZ=48$ and $WQ=72$.

From similarity,

Substitute $WX=88$, $WQ=72$, and $QY=54$:

So $YZ = 88\cdot \frac{3}{4} = 66$.

Therefore, the length of $\overline{YZ}$ is $66$.