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\begin{document}

\sloppy

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

 Dans le triangle GNF rectangle en G, on sait que :
\begin{itemize}
\item GF=8 cm
\item \(\widehat{NFG}\) =  32\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZNG rectangle en Z, on sait que :
\begin{itemize}
\item NG=6,2 cm
\item \(\widehat{NGZ}\) =  22\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZWS rectangle en Z, on sait que :
\begin{itemize}
\item WS=3,6 cm
\item \(\widehat{ZWS}\) = 54\textdegree{}
\end{itemize}

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

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

 Dans le triangle ATK rectangle en A, on sait que :
\begin{itemize}
\item AT=3,3 cm
\item AK=5,1 cm
\end{itemize}

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

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

 Dans le triangle AJW rectangle en A, on sait que :
\begin{itemize}
\item AJ=1,5 cm
\item AW=3,8 cm
\end{itemize}

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

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

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

cos(\(\widehat{GFN}\)) = \(\dfrac{GF}{NF}\)
\par\addvspace{.5em}
  d'où    cos(32\textdegree{}) =  \(\dfrac{8}{NF}\)
\par\addvspace{.5em}
 On a donc NF =  \(\dfrac{8}{\cos(32^{\circ})}\) \(\approx\) 9,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{ZGN}\)) = \(\dfrac{ZG}{NG}\)
\par\addvspace{.5em}
  d'où    cos(22\textdegree{}) =  \(\dfrac{ZG}{6{,}2}\)

On a donc ZG = 6,2 \(\times\) cos(22\textdegree{}) \(\approx\) 5{,}7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{ZWS}\)) = \(\dfrac{ZW}{WS}\)
\par\addvspace{.5em}
d'où cos(54\textdegree{}) = \(\dfrac{ZW}{3{,}6}\)

On a donc ZW = 3,6 \(\times\) cos(54\textdegree{}) \(\approx\) 2,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{AKT}\)) = \(\dfrac{AT}{AK}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{AKT}\)) =  \(\dfrac{3{,}3}{5{,}1}\)
\par\addvspace{.5em}
 On a donc \(\widehat{AKT}\) =  ArcTan(\(\dfrac{3{,}3}{5{,}1}\)) \(\approx\) 33\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{AJW}\)) = \(\dfrac{AW}{AJ}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{AJW}\)) = \(\dfrac{3{,}8}{1{,}5}\)
\par\addvspace{.5em}
 On a donc AJW = ArcTan(\(\dfrac{3{,}8}{1{,}5}\)) \(\approx\) 68\textdegree{}\end{document}
