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\begin{document}

\sloppy

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

 Dans le triangle GNZ rectangle en G, on sait que :
\begin{itemize}
\item GZ=6,4 cm
\item \(\widehat{GNZ}\) = 57\textdegree{}
\end{itemize}

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

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

 Dans le triangle FWL rectangle en F, on sait que :
\begin{itemize}
\item FW=3,6 cm
\item \(\widehat{WLF}\) =  34\textdegree{}
\end{itemize}

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

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

 Dans le triangle VDG rectangle en V, on sait que :
\begin{itemize}
\item DG=5,3 cm
\item \(\widehat{DGV}\) =  40\textdegree{}
\end{itemize}

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

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

 Dans le triangle SPV rectangle en S, on sait que :
\begin{itemize}
\item SP=2,6 cm
\item SV=4,9 cm
\end{itemize}

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

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

 Dans le triangle LND rectangle en L, on sait que :
\begin{itemize}
\item LD=5,4 cm
\item ND=7,1 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{LND}\). \newpage
\section*{\textcolor{red}{Correction}}

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

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

\par\addvspace{.5em}
tan(\(\widehat{GNZ}\)) = \(\dfrac{GZ}{GN}\)
\par\addvspace{.5em}
tan(57\textdegree{}) = \(\dfrac{6{,}4}{GN}\)
\par\addvspace{.5em}
 On a donc GN = \(\dfrac{6{,}4}{\tan(57^{\circ})}\) \(\approx\) 4,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{FLW}\)) = \(\dfrac{FW}{WL}\)
\par\addvspace{.5em}
 D'où sin(34\textdegree{}) = \(\dfrac{3{,}6}{WL}\)
\par\addvspace{.5em}
On a donc WL =\(\dfrac{3{,}6}{\sin(34^{\circ})}\) \(\approx\) 6,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{VGD}\)) = \(\dfrac{VG}{DG}\)
\par\addvspace{.5em}
  d'où    cos(40\textdegree{}) =  \(\dfrac{VG}{5{,}3}\)

On a donc VG = 5,3 \(\times\) cos(40\textdegree{}) \(\approx\) 4{,}1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{SVP}\)) = \(\dfrac{SP}{SV}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{SVP}\)) =  \(\dfrac{2{,}6}{4{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{SVP}\) =  ArcTan(\(\dfrac{2{,}6}{4{,}9}\)) \(\approx\) 28\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.139,2.728) {?};
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 Dans le triangle LND rectangle en L, on cherche une relation entre l'angle aigu   \(\widehat{LND}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{LND}\)) =  \(\dfrac{LD}{ND}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{LND}\)) = \(\dfrac{5{,}4}{7{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{LND}\) =  ArcSin(\(\dfrac{5{,}4}{7{,}1}\)) \(\approx\) 50\textdegree{}\end{document}
