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\begin{document}

\sloppy

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

 Dans le triangle RWJ rectangle en R, on sait que :
\begin{itemize}
\item RW=3,1 cm
\item RJ=4,9 cm
\end{itemize}

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

 Dans le triangle FVA rectangle en F, on sait que :
\begin{itemize}
\item FV=9,7 cm
\item \(\widehat{VAF}\) =  33\textdegree{}
\end{itemize}

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

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

 Dans le triangle DAL rectangle en D, on sait que :
\begin{itemize}
\item AL=6,3 cm
\item \(\widehat{ALD}\) =  31\textdegree{}
\end{itemize}

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

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

 Dans le triangle TZW rectangle en T, on sait que :
\begin{itemize}
\item ZW=7,8 cm
\item \(\widehat{ZWT}\) =  38\textdegree{}
\end{itemize}

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

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

 Dans le triangle JSF rectangle en J, on sait que :
\begin{itemize}
\item JF=6,3 cm
\item SF=8,5 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{RWJ}\)) = \(\dfrac{RJ}{RW}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{RWJ}\)) = \(\dfrac{4{,}9}{3{,}1}\)
\par\addvspace{.5em}
 On a donc RWJ = ArcTan(\(\dfrac{4{,}9}{3{,}1}\)) \(\approx\) 58\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{FAV}\)) = \(\dfrac{FV}{VA}\)
\par\addvspace{.5em}
 D'où sin(33\textdegree{}) = \(\dfrac{9{,}7}{VA}\)
\par\addvspace{.5em}
On a donc VA =\(\dfrac{9{,}7}{\sin(33^{\circ})}\) \(\approx\) 17,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{DLA}\)) = \(\dfrac{DL}{AL}\)
\par\addvspace{.5em}
  d'où    cos(31\textdegree{}) =  \(\dfrac{DL}{6{,}3}\)

On a donc DL = 6,3 \(\times\) cos(31\textdegree{}) \(\approx\) 5{,}4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{TWZ}\)) = \(\dfrac{TZ}{ZW}\)
\par\addvspace{.5em}
 D'où sin(38\textdegree{}) = \(\dfrac{TZ}{7{,}8}\)

On a donc TZ = 7,8 \(\times\) sin(38\textdegree{}) \(\approx\) 4,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

cos(\(\widehat{JFS}\)) = \(\dfrac{JF}{SF}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{JFS}\)) =  \(\dfrac{6{,}3}{8{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{JFS}\) = Arccos (\(\dfrac{6{,}3}{8{,}5}\)) \(\approx\) 42\textdegree{} \end{document}
