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\begin{document}

\sloppy

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

 Dans le triangle ZRN rectangle en Z, on sait que :
\begin{itemize}
\item ZR=8,1 cm
\item \(\widehat{ZRN}\) = 64\textdegree{}
\end{itemize}

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

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

 Dans le triangle AGD rectangle en A, on sait que :
\begin{itemize}
\item AD=4,7 cm
\item GD=9 cm
\end{itemize}

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

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

 Dans le triangle DLV rectangle en D, on sait que :
\begin{itemize}
\item DL=6,8 cm
\item \(\widehat{DLV}\) = 73\textdegree{}
\end{itemize}

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

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

 Dans le triangle VCN rectangle en V, on sait que :
\begin{itemize}
\item CN=2,5 cm
\item \(\widehat{CNV}\) =  36\textdegree{}
\end{itemize}

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

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

 Dans le triangle LVS rectangle en L, on sait que :
\begin{itemize}
\item LS=5,5 cm
\item VS=7,9 cm
\end{itemize}

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{ZRN}\)) = \(\dfrac{ZN}{ZR}\)
\par\addvspace{.5em}
d'où tan(64\textdegree{}) =\(\dfrac{ZN}{8{,}1}\)
\par\addvspace{.5em}
On a donc ZN = 8,1 \(\times\) tan(64\textdegree{}) \(\approx\) 16,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{ADG}\)) = \(\dfrac{AD}{GD}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{ADG}\)) =  \(\dfrac{4{,}7}{9}\)
\par\addvspace{.5em}
On a donc \(\widehat{ADG}\) = Arccos (\(\dfrac{4{,}7}{9}\)) \(\approx\) 59\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{DLV}\)) = \(\dfrac{DL}{LV}\)
\par\addvspace{.5em}
 d'où cos(73\textdegree{}) = \(\dfrac{6{,}8}{LV}\)
\par\addvspace{.5em}
On a donc LV = \(\dfrac{6{,}8}{\cos(73^{\circ})}\)\(\approx\) 23,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{VNC}\)) = \(\dfrac{VC}{CN}\)
\par\addvspace{.5em}
 D'où sin(36\textdegree{}) = \(\dfrac{VC}{2{,}5}\)

On a donc VC = 2,5 \(\times\) sin(36\textdegree{}) \(\approx\) 1,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{LVS}\)) =  \(\dfrac{LS}{VS}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{LVS}\)) = \(\dfrac{5{,}5}{7{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{LVS}\) =  ArcSin(\(\dfrac{5{,}5}{7{,}9}\)) \(\approx\) 44\textdegree{}\end{document}
