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\begin{document}

\sloppy

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

 Dans le triangle GSA rectangle en G, on sait que :
\begin{itemize}
\item GS=2,8 cm
\item GA=4,5 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{GSA}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle FAS rectangle en F, on sait que :
\begin{itemize}
\item FA=9,7 cm
\item \(\widehat{ASF}\) =  13\textdegree{}
\end{itemize}

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

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

 Dans le triangle WHP rectangle en W, on sait que :
\begin{itemize}
\item WP=4,5 cm
\item \(\widehat{WHP}\) = 72\textdegree{}
\end{itemize}

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

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

 Dans le triangle FTP rectangle en F, on sait que :
\begin{itemize}
\item TP=7,3 cm
\item \(\widehat{TPF}\) =  33\textdegree{}
\end{itemize}

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

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

 Dans le triangle JNM rectangle en J, on sait que :
\begin{itemize}
\item JN=3,1 cm
\item JM=3,7 cm
\end{itemize}

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

\newpage
\section*{\textcolor{red}{Correction}}

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

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

\par\addvspace{.5em}
tan(\(\widehat{GSA}\)) = \(\dfrac{GA}{GS}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{GSA}\)) = \(\dfrac{4{,}5}{2{,}8}\)
\par\addvspace{.5em}
 On a donc GSA = ArcTan(\(\dfrac{4{,}5}{2{,}8}\)) \(\approx\) 58\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{FSA}\)) = \(\dfrac{FA}{FS}\)
\par\addvspace{.5em}
d'où  tan(13\textdegree{}) =  \(\dfrac{9{,}7}{FS}\)
\par\addvspace{.5em}
On a donc FS = \(\dfrac{9{,}7}{\tan(13^{\circ})}\) \(\approx\) 42 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{WHP}\)) =  \(\dfrac{WP}{HP}\)
\par\addvspace{.5em}
 d'où sin(72\textdegree{}) = \(\dfrac{4{,}5}{HP}\)
\par\addvspace{.5em}
 On a donc HP = \(\dfrac{4{,}5}{\sin(72^{\circ})}\) \(\approx\) 4,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{FPT}\)) = \(\dfrac{FT}{TP}\)
\par\addvspace{.5em}
 D'où sin(33\textdegree{}) = \(\dfrac{FT}{7{,}3}\)

On a donc FT = 7,3 \(\times\) sin(33\textdegree{}) \(\approx\) 4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {3,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.544,0.761) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{JMN}\)) = \(\dfrac{JN}{JM}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{JMN}\)) =  \(\dfrac{3{,}1}{3{,}7}\)
\par\addvspace{.5em}
 On a donc \(\widehat{JMN}\) =  ArcTan(\(\dfrac{3{,}1}{3{,}7}\)) \(\approx\) 40\textdegree{}\end{document}
