On considère un triangle ABC inscrit dans un cercle $\mathcal{C}$ de rayon $r$.

On note $\widehat{A}=\angle{BAC}$.

\medskip

\begin{tabularx}{\linewidth}{*2{>{\centering\arraybackslash\bfseries}X}}
\begin{center}
\begin{tikzpicture}[scale=0.75]
\draw[red!50!black] (0,0) circle (3cm);
\draw[blue!50!black] (70:3cm) node[above right] {A} -- (200:3cm) node[below left] {B} node[midway,left] {$c$} -- (320:3cm) node[below right] {C} node[midway,below] {$a$} -- (70:3cm) node[midway,right] {$b$};
\draw[green!50!black] (200:3cm) -- (0,0) node[above] {O} -- (320:3cm) node[midway,above] {$r$};
\end{tikzpicture}
\end{center}
&
\begin{center}
\begin{tikzpicture}[scale=0.75]
\draw[red!50!black] (0,0) circle (3cm);
\draw[blue!50!black] (290:3cm) node[below] {A} -- (200:3cm) node[below left] {B} node[midway,below] {$c$} -- (320:3cm) node[below right] {C} node[midway,above] {$a$} -- (290:3cm) node[midway,left] {$b$};
\draw[green!50!black] (200:3cm) -- (0,0) node[above] {O} -- (320:3cm) node[midway,above] {$r$};
\end{tikzpicture}
\end{center}
\\
Cas 1 & Cas 2
\end{tabularx}

\begin{enumerate}
\item Montrer que, dans le cas 1 comme dans le cas 2, on a : $\dfrac{a}{2}=r\sin\widehat{A}$.
\item En déduire que : $\dfrac{a}{\sin\widehat{A}}=\dfrac{b}{\sin\widehat{B}}=\dfrac{c}{\sin\widehat{c}}=2r$.
\item En déduire que l'aire du triangle ABC est : $\mathcal{A}=\dfrac{abc}{4r}$.
\end{enumerate}

