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\begin{document}

\sloppy

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

 Dans le triangle PBD rectangle en P, on sait que :
\begin{itemize}
\item PB=10 cm
\item \(\widehat{BDP}\) =  12\textdegree{}
\end{itemize}

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

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

 Dans le triangle CLN rectangle en C, on sait que :
\begin{itemize}
\item CL=2,8 cm
\item \(\widehat{LNC}\) =  26\textdegree{}
\end{itemize}

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

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

 Dans le triangle CPH rectangle en C, on sait que :
\begin{itemize}
\item CP=2,7 cm
\item PH=8,2 cm
\end{itemize}

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

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

 Dans le triangle HMS rectangle en H, on sait que :
\begin{itemize}
\item HM=1,9 cm
\item MS=8 cm
\end{itemize}

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

 Dans le triangle FGN rectangle en F, on sait que :
\begin{itemize}
\item GN=8,3 cm
\item \(\widehat{GNF}\) =  11\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{PDB}\)) = \(\dfrac{PB}{BD}\)
\par\addvspace{.5em}
 D'où sin(12\textdegree{}) = \(\dfrac{10}{BD}\)
\par\addvspace{.5em}
On a donc BD =\(\dfrac{10}{\sin(12^{\circ})}\) \(\approx\) 48,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{CNL}\)) = \(\dfrac{CL}{CN}\)
\par\addvspace{.5em}
d'où  tan(26\textdegree{}) =  \(\dfrac{2{,}8}{CN}\)
\par\addvspace{.5em}
On a donc CN = \(\dfrac{2{,}8}{\tan(26^{\circ})}\) \(\approx\) 5,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{CHP}\)) = \(\dfrac{CP}{PH}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{CHP}\)) =\(\dfrac{2{,}7}{8{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{CHP}\) = ArcSin(\(\dfrac{2{,}7}{8{,}2}\)) \(\approx\) 19\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{HMS}\)) = \(\dfrac{HM}{MS}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{HMS}\)) = \(\dfrac{1{,}9}{8}\)
\par\addvspace{.5em}
On a donc \(\widehat{HMS}\) =ArcCos(\(\dfrac{1{,}9}{8}\)) \(\approx\)76\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.286,1.521) {11°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{FNG}\)) = \(\dfrac{FG}{GN}\)
\par\addvspace{.5em}
 D'où sin(11\textdegree{}) = \(\dfrac{FG}{8{,}3}\)

On a donc FG = 8,3 \(\times\) sin(11\textdegree{}) \(\approx\) 1,6 cm\end{document}
