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\begin{document}

\sloppy

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

 Dans le triangle JWK rectangle en J, on sait que :
\begin{itemize}
\item WK=2,4 cm
\item \(\widehat{JWK}\) = 47\textdegree{}
\end{itemize}

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

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

 Dans le triangle PGL rectangle en P, on sait que :
\begin{itemize}
\item GL=1,1 cm
\item \(\widehat{GLP}\) =  40\textdegree{}
\end{itemize}

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

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

 Dans le triangle CGB rectangle en C, on sait que :
\begin{itemize}
\item CG=1,4 cm
\item GB=8,5 cm
\end{itemize}

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

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

 Dans le triangle ZCV rectangle en Z, on sait que :
\begin{itemize}
\item ZC=2,4 cm
\item ZV=5,3 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{ZCV}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle SLC rectangle en S, on sait que :
\begin{itemize}
\item SL=7,3 cm
\item \(\widehat{SLC}\) = 58\textdegree{}
\end{itemize}

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

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

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

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{JWK}\)) =  \(\dfrac{JK}{WK}\)
\par\addvspace{.5em}
 d'où sin(47\textdegree{}) = \(\dfrac{JK}{2{,}4}\)

On a donc JK = 2,4 \(\times\) sin(47\textdegree{}) \(\approx\) 1,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{PLG}\)) = \(\dfrac{PG}{GL}\)
\par\addvspace{.5em}
 D'où sin(40\textdegree{}) = \(\dfrac{PG}{1{,}1}\)

On a donc PG = 1,1 \(\times\) sin(40\textdegree{}) \(\approx\) 0,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{ZCV}\)) = \(\dfrac{ZV}{ZC}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{ZCV}\)) = \(\dfrac{5{,}3}{2{,}4}\)
\par\addvspace{.5em}
 On a donc ZCV = ArcTan(\(\dfrac{5{,}3}{2{,}4}\)) \(\approx\) 66\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.583,2.750) {58°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{SLC}\)) = \(\dfrac{SL}{LC}\)
\par\addvspace{.5em}
 d'où cos(58\textdegree{}) = \(\dfrac{7{,}3}{LC}\)
\par\addvspace{.5em}
On a donc LC = \(\dfrac{7{,}3}{\cos(58^{\circ})}\)\(\approx\) 13,8 cm\end{document}
