Answer
The value of a is about $64.4$.
Work Step by Step
At first, we will find the angle $\angle CAD$.
By observing the figure, we get
$\begin{align}
& \angle CAD=\angle EAD-\angle EAC \\
& =32{}^\circ -28{}^\circ \\
& =4{}^\circ
\end{align}$
Now, we will find angle B by the linear sum property:
$\begin{align}
& 35{}^\circ +B=180{}^\circ \\
& B=180{}^\circ -35{}^\circ \\
& B=145{}^\circ
\end{align}$
Now, we will find an angle $\angle ADB$.
Using the sum property of triangles:
$\begin{align}
& \angle BAD+\angle DBA+\angle ADB=180{}^\circ \\
& \angle ADB=180{}^\circ -1{}^\circ -145{}^\circ \\
& \angle ADB=34{}^\circ
\end{align}$
Now by the property of alternate interior angles, we get
$\begin{align}
& \angle ACD=\angle EAC \\
& =28{}^\circ
\end{align}$
Now we will compute the length of side AD using the law of sines in triangle ABD.
$\begin{align}
& \frac{AB}{\sin \left( \angle ADB \right)}=\frac{AD}{\sin \left( \angle ABD \right)} \\
& \frac{450}{\sin \left( 145{}^\circ \right)}=\frac{AD}{\sin \left( 34{}^\circ \right)} \\
& AD=\frac{450\sin \left( 34{}^\circ \right)}{\sin \left( 145{}^\circ \right)}
\end{align}$
Now, to find a we will use of the law of sines in triangle ADC,
$\begin{align}
& \frac{DC}{\sin 4{}^\circ }=\frac{AD}{\sin 28{}^\circ } \\
& \frac{a}{\sin 4{}^\circ }=\frac{450\sin 145{}^\circ }{\sin 34{}^\circ \sin 28{}^\circ } \\
& a=\frac{450\sin 145{}^\circ \sin 4{}^\circ }{\sin 34{}^\circ \sin 28{}^\circ } \\
& a\approx 64.4
\end{align}$
