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\begin{document}

\sloppy

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

 Dans le triangle HJL rectangle en H, on sait que :
\begin{itemize}
\item HL=4 cm
\item JL=7,4 cm
\end{itemize}

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

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

 Dans le triangle SDM rectangle en S, on sait que :
\begin{itemize}
\item DM=7,4 cm
\item \(\widehat{SDM}\) = 75\textdegree{}
\end{itemize}

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

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

 Dans le triangle BDL rectangle en B, on sait que :
\begin{itemize}
\item BD=4,9 cm
\item \(\widehat{BDL}\) = 60\textdegree{}
\end{itemize}

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

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

 Dans le triangle MHS rectangle en M, on sait que :
\begin{itemize}
\item MS=5,5 cm
\item \(\widehat{HSM}\) =  18\textdegree{}
\end{itemize}

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

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

 Dans le triangle CDG rectangle en C, on sait que :
\begin{itemize}
\item CG=6,3 cm
\item DG=7,2 cm
\end{itemize}

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

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

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.814,0.853) {?};
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 Dans le triangle HJL rectangle en H, on cherche une relation entre l'angle aigu   \(\widehat{HLJ}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{HLJ}\)) = \(\dfrac{HL}{JL}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{HLJ}\)) =  \(\dfrac{4}{7{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{HLJ}\) = Arccos (\(\dfrac{4}{7{,}4}\)) \(\approx\) 57\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{SDM}\)) =  \(\dfrac{SM}{DM}\)
\par\addvspace{.5em}
 d'où sin(75\textdegree{}) = \(\dfrac{SM}{7{,}4}\)

On a donc SM = 7,4 \(\times\) sin(75\textdegree{}) \(\approx\) 7,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{BDL}\)) = \(\dfrac{BD}{DL}\)
\par\addvspace{.5em}
 d'où cos(60\textdegree{}) = \(\dfrac{4{,}9}{DL}\)
\par\addvspace{.5em}
On a donc DL = \(\dfrac{4{,}9}{\cos(60^{\circ})}\)\(\approx\) 9,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{MSH}\)) = \(\dfrac{MH}{MS}\)
\par\addvspace{.5em}
d'où  tan(18\textdegree{}) =  \(\dfrac{MH}{5{,}5}\)
\par\addvspace{.5em}
On a donc MH =  5,5 \(\times\) tan(18\textdegree{})  \(\approx\) 1,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {7,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.389,2.715) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{CDG}\)) =  \(\dfrac{CG}{DG}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{CDG}\)) = \(\dfrac{6{,}3}{7{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{CDG}\) =  ArcSin(\(\dfrac{6{,}3}{7{,}2}\)) \(\approx\) 61\textdegree{}\end{document}
