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\begin{document}

\sloppy

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

 Dans le triangle AZH rectangle en A, on sait que :
\begin{itemize}
\item AH=4,9 cm
\item ZH=9,5 cm
\end{itemize}

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

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

 Dans le triangle SWA rectangle en S, on sait que :
\begin{itemize}
\item WA=0,6 cm
\item \(\widehat{WAS}\) =  29\textdegree{}
\end{itemize}

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

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

 Dans le triangle BDH rectangle en B, on sait que :
\begin{itemize}
\item BH=3,1 cm
\item \(\widehat{DHB}\) =  12\textdegree{}
\end{itemize}

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

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

 Dans le triangle CGK rectangle en C, on sait que :
\begin{itemize}
\item CK=1,9 cm
\item \(\widehat{CGK}\) = 51\textdegree{}
\end{itemize}

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

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

 Dans le triangle AMJ rectangle en A, on sait que :
\begin{itemize}
\item AM=3,2 cm
\item MJ=7,4 cm
\end{itemize}

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

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

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

cos(\(\widehat{AHZ}\)) = \(\dfrac{AH}{ZH}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{AHZ}\)) =  \(\dfrac{4{,}9}{9{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{AHZ}\) = Arccos (\(\dfrac{4{,}9}{9{,}5}\)) \(\approx\) 59\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{SAW}\)) = \(\dfrac{SA}{WA}\)
\par\addvspace{.5em}
  d'où    cos(29\textdegree{}) =  \(\dfrac{SA}{0{,}6}\)

On a donc SA = 0,6 \(\times\) cos(29\textdegree{}) \(\approx\) 0{,}5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{BHD}\)) = \(\dfrac{BH}{DH}\)
\par\addvspace{.5em}
  d'où    cos(12\textdegree{}) =  \(\dfrac{3{,}1}{DH}\)
\par\addvspace{.5em}
 On a donc DH =  \(\dfrac{3{,}1}{\cos(12^{\circ})}\) \(\approx\) 3,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{CGK}\)) = \(\dfrac{CK}{CG}\)
\par\addvspace{.5em}
tan(51\textdegree{}) = \(\dfrac{1{,}9}{CG}\)
\par\addvspace{.5em}
 On a donc CG = \(\dfrac{1{,}9}{\tan(51^{\circ})}\) \(\approx\) 1,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{AMJ}\)) = \(\dfrac{AM}{MJ}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{AMJ}\)) = \(\dfrac{3{,}2}{7{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{AMJ}\) =ArcCos(\(\dfrac{3{,}2}{7{,}4}\)) \(\approx\)64\textdegree{}\end{document}
