ABCDE is a regular pentagon and a rectangle AXYE is drawn on the side AE such that the vertices X and Y lie on the sides BC and CD respectively. Calculate t...

(i) Interior angle of the pentagon. The sum of the interior angles of an \(n\)-sided polygon is \((n - 2)\times 180^\circ\). For a regular pentagon \(n = 5\):

\[\text{Each interior angle} = \frac{(5 - 2)\times 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ.\]

(ii) < BXA. In the rectangle \(AXYE\), the side \(AE\) is a side of the pentagon and \(AX \perp AE\), so \(< XAE = 90^\circ\).

The interior angle of the pentagon at \(A\) is \(< BAE = 108^\circ\). Since \(X\) lies on \(BC\), the ray \(AX\) splits \(< BAE\):

\[< BAX = < BAE - < XAE = 108^\circ - 90^\circ = 18^\circ.\]

Also \(X\) lies on \(BC\), so in triangle \(ABX\) the angle at \(B\) equals the interior angle of the pentagon:

\[< ABX = 108^\circ.\]

The angles of triangle \(ABX\) sum to \(180^\circ\):

\[< BXA = 180^\circ - < ABX - < BAX = 180^\circ - 108^\circ - 18^\circ = 54^\circ.\]

Therefore \(< BXA = \mathbf{54^\circ}\).