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\begin{document}

\sloppy

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

 Dans le triangle CGJ rectangle en C, on sait que :
\begin{itemize}
\item CG=1,1 cm
\item GJ=9,2 cm
\end{itemize}

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

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

 Dans le triangle CRW rectangle en C, on sait que :
\begin{itemize}
\item RW=7 cm
\item \(\widehat{RWC}\) =  40\textdegree{}
\end{itemize}

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

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

 Dans le triangle WGF rectangle en W, on sait que :
\begin{itemize}
\item GF=5,5 cm
\item \(\widehat{GFW}\) =  41\textdegree{}
\end{itemize}

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

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

 Dans le triangle NVA rectangle en N, on sait que :
\begin{itemize}
\item NA=5,5 cm
\item VA=9,1 cm
\end{itemize}

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

 Dans le triangle LKW rectangle en L, on sait que :
\begin{itemize}
\item LK=7,8 cm
\item \(\widehat{LKW}\) = 54\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{CJG}\)) = \(\dfrac{CG}{GJ}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{CJG}\)) =\(\dfrac{1{,}1}{9{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{CJG}\) = ArcSin(\(\dfrac{1{,}1}{9{,}2}\)) \(\approx\) 7\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{CWR}\)) = \(\dfrac{CR}{RW}\)
\par\addvspace{.5em}
 D'où sin(40\textdegree{}) = \(\dfrac{CR}{7}\)

On a donc CR = 7 \(\times\) sin(40\textdegree{}) \(\approx\) 4,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{WFG}\)) = \(\dfrac{WF}{GF}\)
\par\addvspace{.5em}
  d'où    cos(41\textdegree{}) =  \(\dfrac{WF}{5{,}5}\)

On a donc WF = 5,5 \(\times\) cos(41\textdegree{}) \(\approx\) 4{,}2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{NVA}\)) =  \(\dfrac{NA}{VA}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{NVA}\)) = \(\dfrac{5{,}5}{9{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{NVA}\) =  ArcSin(\(\dfrac{5{,}5}{9{,}1}\)) \(\approx\) 37\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{LKW}\)) = \(\dfrac{LK}{KW}\)
\par\addvspace{.5em}
 d'où cos(54\textdegree{}) = \(\dfrac{7{,}8}{KW}\)
\par\addvspace{.5em}
On a donc KW = \(\dfrac{7{,}8}{\cos(54^{\circ})}\)\(\approx\) 13,3 cm\end{document}
