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\begin{document}

\sloppy

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

 Dans le triangle VWF rectangle en V, on sait que :
\begin{itemize}
\item VW=4,4 cm
\item \(\widehat{WFV}\) =  13\textdegree{}
\end{itemize}

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

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

 Dans le triangle JLC rectangle en J, on sait que :
\begin{itemize}
\item LC=5,6 cm
\item \(\widehat{LCJ}\) =  14\textdegree{}
\end{itemize}

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

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

 Dans le triangle TGN rectangle en T, on sait que :
\begin{itemize}
\item TN=4,2 cm
\item GN=8,1 cm
\end{itemize}

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

 Dans le triangle NZD rectangle en N, on sait que :
\begin{itemize}
\item ZD=6,1 cm
\item \(\widehat{ZDN}\) =  39\textdegree{}
\end{itemize}

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

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

 Dans le triangle KFW rectangle en K, on sait que :
\begin{itemize}
\item KF=2,6 cm
\item FW=9,1 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{VFW}\)) = \(\dfrac{VW}{WF}\)
\par\addvspace{.5em}
 D'où sin(13\textdegree{}) = \(\dfrac{4{,}4}{WF}\)
\par\addvspace{.5em}
On a donc WF =\(\dfrac{4{,}4}{\sin(13^{\circ})}\) \(\approx\) 19,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{JCL}\)) = \(\dfrac{JL}{LC}\)
\par\addvspace{.5em}
 D'où sin(14\textdegree{}) = \(\dfrac{JL}{5{,}6}\)

On a donc JL = 5,6 \(\times\) sin(14\textdegree{}) \(\approx\) 1,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{TGN}\)) =  \(\dfrac{TN}{GN}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{TGN}\)) = \(\dfrac{4{,}2}{8{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{TGN}\) =  ArcSin(\(\dfrac{4{,}2}{8{,}1}\)) \(\approx\) 31\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{NDZ}\)) = \(\dfrac{ND}{ZD}\)
\par\addvspace{.5em}
  d'où    cos(39\textdegree{}) =  \(\dfrac{ND}{6{,}1}\)

On a donc ND = 6,1 \(\times\) cos(39\textdegree{}) \(\approx\) 4{,}7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{KWF}\)) = \(\dfrac{KF}{FW}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{KWF}\)) =\(\dfrac{2{,}6}{9{,}1}\)
\par\addvspace{.5em}
 On a donc \(\widehat{KWF}\) = ArcSin(\(\dfrac{2{,}6}{9{,}1}\)) \(\approx\) 17\textdegree{}\end{document}
