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\begin{document}

\sloppy

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

 Dans le triangle CZG rectangle en C, on sait que :
\begin{itemize}
\item CZ=1,9 cm
\item \(\widehat{ZGC}\) =  38\textdegree{}
\end{itemize}

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

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

 Dans le triangle KSL rectangle en K, on sait que :
\begin{itemize}
\item KL=4 cm
\item SL=9,8 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{KSL}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle MAV rectangle en M, on sait que :
\begin{itemize}
\item MA=7 cm
\item \(\widehat{AVM}\) =  35\textdegree{}
\end{itemize}

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

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

 Dans le triangle AVG rectangle en A, on sait que :
\begin{itemize}
\item AG=1,8 cm
\item \(\widehat{AVG}\) = 77\textdegree{}
\end{itemize}

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

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

 Dans le triangle GKN rectangle en G, on sait que :
\begin{itemize}
\item GK=1,5 cm
\item GN=6 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{CGZ}\)) = \(\dfrac{CZ}{CG}\)
\par\addvspace{.5em}
d'où  tan(38\textdegree{}) =  \(\dfrac{1{,}9}{CG}\)
\par\addvspace{.5em}
On a donc CG = \(\dfrac{1{,}9}{\tan(38^{\circ})}\) \(\approx\) 2,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{KSL}\)) =  \(\dfrac{KL}{SL}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{KSL}\)) = \(\dfrac{4}{9{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{KSL}\) =  ArcSin(\(\dfrac{4}{9{,}8}\)) \(\approx\) 24\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{MVA}\)) = \(\dfrac{MA}{AV}\)
\par\addvspace{.5em}
 D'où sin(35\textdegree{}) = \(\dfrac{7}{AV}\)
\par\addvspace{.5em}
On a donc AV =\(\dfrac{7}{\sin(35^{\circ})}\) \(\approx\) 12,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{AVG}\)) = \(\dfrac{AG}{AV}\)
\par\addvspace{.5em}
tan(77\textdegree{}) = \(\dfrac{1{,}8}{AV}\)
\par\addvspace{.5em}
 On a donc AV = \(\dfrac{1{,}8}{\tan(77^{\circ})}\) \(\approx\) 0,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.288,1.402) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{GNK}\)) = \(\dfrac{GK}{GN}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{GNK}\)) =  \(\dfrac{1{,}5}{6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{GNK}\) =  ArcTan(\(\dfrac{1{,}5}{6}\)) \(\approx\) 14\textdegree{}\end{document}
