## Precalculus (6th Edition) Blitzer

The value of a is about $64.4$.
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}