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\begin{document}

\sloppy

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

 Dans le triangle DLP rectangle en D, on sait que :
\begin{itemize}
\item DP=5,1 cm
\item LP=9,2 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{DLP}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle GHN rectangle en G, on sait que :
\begin{itemize}
\item GN=9,6 cm
\item \(\widehat{GHN}\) = 70\textdegree{}
\end{itemize}

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

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

 Dans le triangle BLF rectangle en B, on sait que :
\begin{itemize}
\item BL=2,8 cm
\item BF=3,5 cm
\end{itemize}

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

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

 Dans le triangle RFG rectangle en R, on sait que :
\begin{itemize}
\item FG=2,6 cm
\item \(\widehat{RFG}\) = 51\textdegree{}
\end{itemize}

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

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

 Dans le triangle DAM rectangle en D, on sait que :
\begin{itemize}
\item DM=7,5 cm
\item \(\widehat{AMD}\) =  28\textdegree{}
\end{itemize}

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

\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 (3.667,2.000) {9,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.761,2.695) {?};
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 Dans le triangle DLP rectangle en D, on cherche une relation entre l'angle aigu   \(\widehat{DLP}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{DLP}\)) =  \(\dfrac{DP}{LP}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{DLP}\)) = \(\dfrac{5{,}1}{9{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{DLP}\) =  ArcSin(\(\dfrac{5{,}1}{9{,}2}\)) \(\approx\) 34\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{GHN}\)) = \(\dfrac{GN}{GH}\)
\par\addvspace{.5em}
tan(70\textdegree{}) = \(\dfrac{9{,}6}{GH}\)
\par\addvspace{.5em}
 On a donc GH = \(\dfrac{9{,}6}{\tan(70^{\circ})}\) \(\approx\) 3,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{BFL}\)) = \(\dfrac{BL}{BF}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{BFL}\)) =  \(\dfrac{2{,}8}{3{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{BFL}\) =  ArcTan(\(\dfrac{2{,}8}{3{,}5}\)) \(\approx\) 39\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{RFG}\)) =  \(\dfrac{RG}{FG}\)
\par\addvspace{.5em}
 d'où sin(51\textdegree{}) = \(\dfrac{RG}{2{,}6}\)

On a donc RG = 2,6 \(\times\) sin(51\textdegree{}) \(\approx\) 2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{DMA}\)) = \(\dfrac{DM}{AM}\)
\par\addvspace{.5em}
  d'où    cos(28\textdegree{}) =  \(\dfrac{7{,}5}{AM}\)
\par\addvspace{.5em}
 On a donc AM =  \(\dfrac{7{,}5}{\cos(28^{\circ})}\) \(\approx\) 8,5 cm\end{document}
