ABCDE is a regular pentagon and DEFG is a square. Find the measures of $\angle \mathbf{EAF}$, $\angle \mathbf{AFD}$and $\angle \mathbf{DAF}$.

As$ABCDE$is a regular pentagon and$DEFG$is a square then,$AB=BC=CD=DE=AE=EF=FG=DG$and all the interior angles of a regular pentagon is 108 degrees then, $\angle ABC=\angle BCD=\angle CDE=\angle DEA=\angle EAB=108°$.

All angles of a square are 90 degrees then, localid="1649848898958" $\angle DEF=\angle EFG=\angle FGD=\angle GDA=90°$.

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

Then,

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

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

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

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

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

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

Then by an angle sum theorem,

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

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

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

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