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\begin{document}

\sloppy

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

 Dans le triangle HGM rectangle en H, on sait que :
\begin{itemize}
\item HG=8,5 cm
\item \(\widehat{HGM}\) = 77\textdegree{}
\end{itemize}

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

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

 Dans le triangle ASJ rectangle en A, on sait que :
\begin{itemize}
\item AS=2,3 cm
\item AJ=4,2 cm
\end{itemize}

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

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

 Dans le triangle HZP rectangle en H, on sait que :
\begin{itemize}
\item HP=4,4 cm
\item \(\widehat{HZP}\) = 77\textdegree{}
\end{itemize}

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

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

 Dans le triangle HAS rectangle en H, on sait que :
\begin{itemize}
\item HS=4 cm
\item AS=6,9 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{HAS}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle GSC rectangle en G, on sait que :
\begin{itemize}
\item GC=4,6 cm
\item \(\widehat{GSC}\) = 73\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{HGM}\)) = \(\dfrac{HM}{HG}\)
\par\addvspace{.5em}
d'où tan(77\textdegree{}) =\(\dfrac{HM}{8{,}5}\)
\par\addvspace{.5em}
On a donc HM = 8,5 \(\times\) tan(77\textdegree{}) \(\approx\) 36,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{AJS}\)) = \(\dfrac{AS}{AJ}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{AJS}\)) =  \(\dfrac{2{,}3}{4{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{AJS}\) =  ArcTan(\(\dfrac{2{,}3}{4{,}2}\)) \(\approx\) 29\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{HZP}\)) = \(\dfrac{HP}{HZ}\)
\par\addvspace{.5em}
tan(77\textdegree{}) = \(\dfrac{4{,}4}{HZ}\)
\par\addvspace{.5em}
 On a donc HZ = \(\dfrac{4{,}4}{\tan(77^{\circ})}\) \(\approx\) 1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{HAS}\)) =  \(\dfrac{HS}{AS}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{HAS}\)) = \(\dfrac{4}{6{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{HAS}\) =  ArcSin(\(\dfrac{4}{6{,}9}\)) \(\approx\) 35\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.447,2.146) {73°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{GSC}\)) =  \(\dfrac{GC}{SC}\)
\par\addvspace{.5em}
 d'où sin(73\textdegree{}) = \(\dfrac{4{,}6}{SC}\)
\par\addvspace{.5em}
 On a donc SC = \(\dfrac{4{,}6}{\sin(73^{\circ})}\) \(\approx\) 4,8 cm\end{document}
