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\begin{document}

\sloppy

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

 Dans le triangle JFL rectangle en J, on sait que :
\begin{itemize}
\item FL=3,1 cm
\item \(\widehat{JFL}\) = 66\textdegree{}
\end{itemize}

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

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

 Dans le triangle TNM rectangle en T, on sait que :
\begin{itemize}
\item TM=6,3 cm
\item \(\widehat{TNM}\) = 48\textdegree{}
\end{itemize}

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

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

 Dans le triangle SBT rectangle en S, on sait que :
\begin{itemize}
\item ST=5,3 cm
\item \(\widehat{BTS}\) =  36\textdegree{}
\end{itemize}

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

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

 Dans le triangle CBA rectangle en C, on sait que :
\begin{itemize}
\item CB=1,4 cm
\item CA=5,2 cm
\end{itemize}

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

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

 Dans le triangle WPS rectangle en W, on sait que :
\begin{itemize}
\item WS=6,2 cm
\item PS=6,8 cm
\end{itemize}

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{JFL}\)) =  \(\dfrac{JL}{FL}\)
\par\addvspace{.5em}
 d'où sin(66\textdegree{}) = \(\dfrac{JL}{3{,}1}\)

On a donc JL = 3,1 \(\times\) sin(66\textdegree{}) \(\approx\) 2,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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 Dans le triangle TNM rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TNM}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{TNM}\)) =  \(\dfrac{TM}{NM}\)
\par\addvspace{.5em}
 d'où sin(48\textdegree{}) = \(\dfrac{6{,}3}{NM}\)
\par\addvspace{.5em}
 On a donc NM = \(\dfrac{6{,}3}{\sin(48^{\circ})}\) \(\approx\) 8,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{STB}\)) = \(\dfrac{SB}{ST}\)
\par\addvspace{.5em}
d'où  tan(36\textdegree{}) =  \(\dfrac{SB}{5{,}3}\)
\par\addvspace{.5em}
On a donc SB =  5,3 \(\times\) tan(36\textdegree{})  \(\approx\) 3,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{CAB}\)) = \(\dfrac{CB}{CA}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{CAB}\)) =  \(\dfrac{1{,}4}{5{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{CAB}\) =  ArcTan(\(\dfrac{1{,}4}{5{,}2}\)) \(\approx\) 15\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.412,2.477) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{WPS}\)) =  \(\dfrac{WS}{PS}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{WPS}\)) = \(\dfrac{6{,}2}{6{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{WPS}\) =  ArcSin(\(\dfrac{6{,}2}{6{,}8}\)) \(\approx\) 66\textdegree{}\end{document}
