In the figure, WXYZ is a trapezium and AXYB is a rhombus. XBC and XAZ are straight lines. WX // ZY, BC = BY, $\angle XAB = 82^\circ$, $\angle ZYB = 19^\circ$ and $\angle WZX = 57^\circ$.

1. Find the interior angle of the rhombus at B

Since $AXYB$ is a rhombus, adjacent angles are supplementary.
$\angle YBA = 180^\circ - \angle XAB = 180^\circ - 82^\circ = 98^\circ$

2. Find the angle $\angle YBX$ using the diagonal property of a rhombus

The diagonal $XB$ bisects the interior angle at $B$.
$\angle YBX = \frac{98^\circ}{2} = 49^\circ$

3. Find $\angle YBC$ on the straight line $XBC$

Since $XBC$ is a straight line, angles on a straight line add up to $180^\circ$.
$\angle YBC = 180^\circ - \angle YBX = 180^\circ - 49^\circ = 131^\circ$

4. Calculate $\angle c$ using the isosceles triangle $YBC$

In $\triangle YBC$, we are given $BC = BY$, so it is an isosceles triangle.
$\angle c = \angle BCY = \frac{180^\circ - 131^\circ}{2} = \frac{49^\circ}{2} = 24.5^\circ$

1. Identify parallel lines from the rhombus to find $\angle XZY$.

Since $AXYB$ is a rhombus, its opposite sides are parallel, so $AX \parallel BY$.
Since $XAZ$ is a straight line, the line $XZ$ is parallel to $BY$.
Therefore, alternate interior angles are equal:
$\angle XZY = \angle ZYB = 19^\circ$

2. Find $\angle WZY$ and use the trapezium's parallel sides to find $\angle w$.

$\angle WZY = \angle WZX + \angle XZY = 57^\circ + 19^\circ = 76^\circ$
In the trapezium $WXYZ$, we are given $WX \parallel ZY$.
Interior angles on the same side of the transversal $WZ$ sum to $180^\circ$:
$\angle w + \angle WZY = 180^\circ$
$\angle w = 180^\circ - 76^\circ = 104^\circ$