The diagram shows a rhombus WXYZ, where the diagonals intersect at D. ZDEX lies on a straight line. EY = 4.7 cm, XZ = 15 cm and $\angle DYE = 42^\circ$. Find the length EX.

1. Find $DX$ using the fact that diagonals of a rhombus bisect each other.

Since the diagonals of a rhombus bisect each other,
$DX=\frac{XZ}{2}=\frac{15}{2}=7.5\text{ cm}.$

2. Find $DE$ using trigonometry in right-angled triangle $DYE$.

The diagonals of a rhombus are perpendicular, so $\angle YDE=90^\circ$.

In right-angled triangle $DYE$,
$\sin 42^\circ=\frac{DE}{EY}.$
Hence
$DE=4.7\sin 42^\circ=3.1449\ldots\text{ cm}.$

3. Subtract $DE$ from $DX$ to find $EX$.

Since $D$, $E$ and $X$ are on the same straight line with $E$ between $D$ and $X$,
$EX=DX-DE$
$=7.5-3.1449\ldots$
$=4.3550\ldots\text{ cm}.$