\begin{enumerate}
\item 
\begin{enumerate}
\item $\dfrac{\numprint{2015}\pi}{7}$.

On calcule : $\numprint{2015}\div7\approx287,85\approx288$ donc $\numprint{2015}=288\times7-1$ d'où :
\begin{align*}
\dfrac{\numprint{2015}\pi}{7} & = \dfrac{(288\times7-1)\pi}{7}\\
 & =288\pi-\dfrac{\pi}{7}\\
\Aboxed{\dfrac{\numprint{2015}\pi}{7} & =-\dfrac{\pi}{7}}
\end{align*}

\item $\dfrac{\numprint{1695}\pi}{9}$.

On a : $\numprint{1695}\div9\approx188,33\approx188$, donc on écrit : $\numprint{1695}=188\times9+2$ et donc :
\begin{align*}
\dfrac{\numprint{1695}\pi}{9} & = \dfrac{(188\times9+2)\pi}{9}\\
& = 188\pi+\dfrac{2\pi}{9}\\
\Aboxed{\dfrac{\numprint{1695}\pi}{9} & = \dfrac{2\pi}{9}}
\end{align*}

\item $\dfrac{\numprint{2001}\pi}{4}$.

On a : $\numprint{2001}\div4=500,25=500\times4+1$ donc :
\begin{align*}
\dfrac{\numprint{2001}\pi}{4} & = 500\pi+\dfrac{1\pi}{4}\\
\Aboxed{\dfrac{\numprint{2001}\pi}{4} & = \dfrac{\pi}{4}}
\end{align*}
\end{enumerate}

\item
\begin{itemize}
\item  $\begin{aligned}[t]
\angleor{ED}{EB} & = \angleor{ED}{EA}+\angleor{EA}{EB}\\
& = \underbrace{\dfrac{1}{2}\left(\pi-\dfrac{\pi}{6}\right)}_{\text{DAE isocèle en A}}+\dfrac{\pi}{3}\\
& = \dfrac{5\pi}{12}+\dfrac{4\pi}{12}\\
& = \dfrac{9\pi}{12}\\
\Aboxed{\angleor{ED}{EB} & = \dfrac{3\pi}{4}}
\end{aligned}$

\item  $\begin{aligned}[t]
\angleor{EB}{EF} & = \dfrac{1}{2}\left(\pi-\dfrac{\pi}{6}-\dfrac{\pi}{3}\right)\qquad\text{(EBF isocèle en B)}\\
\Aboxed{\angleor{EB}{EF} & = \dfrac{\pi}{4}}
\end{aligned}$

\item $\begin{aligned}[t]
\angleor{ED}{EF} & = \angleor{ED}{EB}+\angleor{EB}{EF}\\
& = \dfrac{3\pi}{4}+\dfrac{\pi}{4}\\
\Aboxed{\angleor{ED}{EF} & = \pi}
\end{aligned}$

On en conclut que D, E et F sont alignés.
\end{itemize}
\end{enumerate}