$\mathit{A}\mathit{B}\mathit{C}\mathit{D}\mathit{E}$is a regular pentagon and $\mathit{D}\mathit{E}\mathit{F}\mathit{G}$is a square. Find the measures of $\mathbf{\angle }\mathbf{EAF}\mathbf{,}\mathbf{}\mathbf{\angle }\mathbf{AFD}$ and $\angle \mathbf{DAF}$.

As$ABCDE$ is a regular pentagon and $DEFG$is a square then, and all the interior angles of a regular pentagon is 108 degrees then, role="math" localid="1648817128078" $AB=BC=CD=DE=AE=EF=FG=DG$.

All angles of a square are 90 degrees then $\angle ABC=\angle BCD=\angle CDE=\angle DEA=\angle EAB=108°$, $\angle DEF=\angle EFG=\angle FGD=\angle GDA=90°$.

Consider$\Delta AEF$ in which $AE=EF$, therefore$\Delta AEF$ is an isosceles triangle, then $m\angle EAF=m\angle EFA$.

Then,

$\begin{array}{l}m\angle AEF+m\angle AED+m\angle FED=360\\ m\angle AEF+108+90=360\\ m\angle AEF=360-198\\ =162\end{array}$

Consider$\Delta AEF$ then by an angle sum theorem,

$\begin{array}{l}m\angle AEF+m\angle EFA+m\angle EAF=180\\ 162+2m\angle EAF=180\\ 2m\angle EAF=18\\ m\angle EAF=9°\end{array}$

$\angle AFD$can be written as$\angle AFD=\angle EFA+EFD$ then

$\begin{array}{l}m\angle AFD=m\angle EFA+m\angle EFD\\ =9+45\\ =54°\end{array}$

Consider $\Delta AED$in which $AE=ED$, therefore $\Delta AED$is an isosceles triangle, then $m\angle DAE=m\angle EDA$.

Then by an angle sum theorem,

$\begin{array}{l}m\angle DAE+m\angle EDA+m\angle AED=180\\ 2m\angle DAE+108=180\\ 2m\angle DAE=72\\ m\angle DAE=36°\end{array}$

$\angle DAF$can be written as $\angle DAF=\angle DAE+\angle EAF$then

$\begin{array}{l}m\angle DAF=m\angle DAE+m\angle EAF\\ =36+9\\ =45°\end{array}$

Therefore, the measures of $\angle EAF,\angle AFD$and $\angle DAF$is $9°,54°$and $45°$respectively.