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\begin{document}

\sloppy

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

 Dans le triangle ZSW rectangle en Z, on sait que :
\begin{itemize}
\item ZS=7,7 cm
\item \(\widehat{SWZ}\) =  19\textdegree{}
\end{itemize}

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

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

 Dans le triangle LBS rectangle en L, on sait que :
\begin{itemize}
\item LB=5,5 cm
\item \(\widehat{LBS}\) = 54\textdegree{}
\end{itemize}

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

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

 Dans le triangle GJL rectangle en G, on sait que :
\begin{itemize}
\item GL=3,6 cm
\item JL=8,6 cm
\end{itemize}

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

 Dans le triangle RAS rectangle en R, on sait que :
\begin{itemize}
\item AS=9,8 cm
\item \(\widehat{ASR}\) =  32\textdegree{}
\end{itemize}

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

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

 Dans le triangle DGR rectangle en D, on sait que :
\begin{itemize}
\item DG=1 cm
\item DR=5,6 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{ZWS}\)) = \(\dfrac{ZS}{ZW}\)
\par\addvspace{.5em}
d'où  tan(19\textdegree{}) =  \(\dfrac{7{,}7}{ZW}\)
\par\addvspace{.5em}
On a donc ZW = \(\dfrac{7{,}7}{\tan(19^{\circ})}\) \(\approx\) 22,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{LBS}\)) = \(\dfrac{LB}{BS}\)
\par\addvspace{.5em}
 d'où cos(54\textdegree{}) = \(\dfrac{5{,}5}{BS}\)
\par\addvspace{.5em}
On a donc BS = \(\dfrac{5{,}5}{\cos(54^{\circ})}\)\(\approx\) 9,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{GJL}\)) =  \(\dfrac{GL}{JL}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{GJL}\)) = \(\dfrac{3{,}6}{8{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{GJL}\) =  ArcSin(\(\dfrac{3{,}6}{8{,}6}\)) \(\approx\) 25\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{RSA}\)) = \(\dfrac{RA}{AS}\)
\par\addvspace{.5em}
 D'où sin(32\textdegree{}) = \(\dfrac{RA}{9{,}8}\)

On a donc RA = 9,8 \(\times\) sin(32\textdegree{}) \(\approx\) 5,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
tan(\(\widehat{DRG}\)) = \(\dfrac{DG}{DR}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{DRG}\)) =  \(\dfrac{1}{5{,}6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{DRG}\) =  ArcTan(\(\dfrac{1}{5{,}6}\)) \(\approx\) 10\textdegree{}\end{document}
