We are given a kite $WXYZ$. We are given $m\angle WXY = 120^\circ$, $m\angle WZY = (4x)^\circ$, and $m\angle ZWX = (10x)^\circ$. We need to find $m\angle ZYX$.

1. Problem Description

We are given a kite $WXYZ$. We are given $m\angle WXY = 120^\circ$, $m\angle WZY = (4x)^\circ$, and $m\angle ZWX = (10x)^\circ$. We need to find $m\angle ZYX$.

2. Solution Steps

In a kite, two pairs of adjacent sides are equal in length. Also, one pair of opposite angles are equal. The angles between the unequal sides are equal. In kite $WXYZ$, the angles $WXY$ and $WZY$ are not necessarily equal, so the other two angles $ZWX$ and $ZYX$ are equal.

Therefore, $m\angle ZYX = m\angle ZWX$.

The sum of the interior angles of a quadrilateral is $360^\circ$. Therefore, in kite $WXYZ$, we have

$$m\angle WXY + m\angle WZY + m\angle ZWX + m\angle ZYX = 360^\circ$$

Substituting the given values, we have

$$120^\circ + (4x)^\circ + (10x)^\circ + m\angle ZYX = 360^\circ$$

Since $m\angle ZYX = m\angle ZWX = (10x)^\circ$, we have

$$120 + 4x + 10x + 10x = 360$$

$$120 + 24x = 360$$

$$24x = 360 - 120$$

$$24x = 240$$

$$x = \frac{240}{24}$$

$$x = 10$$

Therefore, $m\angle ZYX = 10x = 10(10) = 100$.

3. Final Answer

100