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\begin{document}

\sloppy

\needspace{4\baselineskip}
\section*{
Exercice 1}

 Dans le triangle GDC rectangle en G, on sait que :
\begin{itemize}
\item GD=7,1 cm
\item \(\widehat{DCG}\) =  22\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [GC]. (Arrondir au dixième)

\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle JSN rectangle en J, on sait que :
\begin{itemize}
\item SN=4,4 cm
\item \(\widehat{SNJ}\) =  29\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [JS]. (Arrondir au dixième)

\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle GAC rectangle en G, on sait que :
\begin{itemize}
\item GA=1,8 cm
\item AC=9,1 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{GCA}\).

\needspace{4\baselineskip}
\section*{
Exercice 4}

 Dans le triangle RFW rectangle en R, on sait que :
\begin{itemize}
\item RW=2,7 cm
\item \(\widehat{FWR}\) =  35\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [WF]. (Arrondir au dixième)

\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle MAS rectangle en M, on sait que :
\begin{itemize}
\item MA=2,2 cm
\item AS=9,4 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{MAS}\). \newpage
\section*{\textcolor{red}{Correction}}

\needspace{4\baselineskip}
\subsection*{
Exercice 1}

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 Dans le triangle GDC rectangle en G, on cherche une relation entre l'angle aigu   \(\widehat{GCD}\) son coté opposé et son coté adjacent.

\par\addvspace{.5em}
tan(\(\widehat{GCD}\)) = \(\dfrac{GD}{GC}\)
\par\addvspace{.5em}
d'où  tan(22\textdegree{}) =  \(\dfrac{7{,}1}{GC}\)
\par\addvspace{.5em}
On a donc GC = \(\dfrac{7{,}1}{\tan(22^{\circ})}\) \(\approx\) 17,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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 Dans le triangle JSN rectangle en J, on cherche une relation entre l'angle aigu   \(\widehat{JNS}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{JNS}\)) = \(\dfrac{JS}{SN}\)
\par\addvspace{.5em}
 D'où sin(29\textdegree{}) = \(\dfrac{JS}{4{,}4}\)

On a donc JS = 4,4 \(\times\) sin(29\textdegree{}) \(\approx\) 2,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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 Dans le triangle GAC rectangle en G, on cherche une relation entre l'angle aigu   \(\widehat{GCA}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{GCA}\)) = \(\dfrac{GA}{AC}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{GCA}\)) =\(\dfrac{1{,}8}{9{,}1}\)
\par\addvspace{.5em}
 On a donc \(\widehat{GCA}\) = ArcSin(\(\dfrac{1{,}8}{9{,}1}\)) \(\approx\) 11\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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 Dans le triangle RFW rectangle en R, on cherche une relation entre l'angle aigu   \(\widehat{RWF}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{RWF}\)) = \(\dfrac{RW}{FW}\)
\par\addvspace{.5em}
  d'où    cos(35\textdegree{}) =  \(\dfrac{2{,}7}{FW}\)
\par\addvspace{.5em}
 On a donc FW =  \(\dfrac{2{,}7}{\cos(35^{\circ})}\) \(\approx\) 3,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.463,1.999) {?};
\end{tikzpicture}\end{center}


 Dans le triangle MAS rectangle en M, on cherche une relation entre l'angle aigu   \(\widehat{MAS}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{MAS}\)) = \(\dfrac{MA}{AS}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{MAS}\)) = \(\dfrac{2{,}2}{9{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{MAS}\) =ArcCos(\(\dfrac{2{,}2}{9{,}4}\)) \(\approx\)76\textdegree{}\end{document}
