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\begin{document}

\sloppy

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

 Dans le triangle HCL rectangle en H, on sait que :
\begin{itemize}
\item HL=9,2 cm
\item \(\widehat{HCL}\) = 57\textdegree{}
\end{itemize}

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

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

 Dans le triangle JSH rectangle en J, on sait que :
\begin{itemize}
\item JH=0,9 cm
\item \(\widehat{JSH}\) = 55\textdegree{}
\end{itemize}

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

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

 Dans le triangle NDC rectangle en N, on sait que :
\begin{itemize}
\item ND=1,9 cm
\item DC=7,1 cm
\end{itemize}

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

 Dans le triangle WNP rectangle en W, on sait que :
\begin{itemize}
\item WN=8,6 cm
\item \(\widehat{NPW}\) =  25\textdegree{}
\end{itemize}

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

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

 Dans le triangle ANS rectangle en A, on sait que :
\begin{itemize}
\item AN=3 cm
\item AS=3,6 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{HCL}\)) =  \(\dfrac{HL}{CL}\)
\par\addvspace{.5em}
 d'où sin(57\textdegree{}) = \(\dfrac{9{,}2}{CL}\)
\par\addvspace{.5em}
 On a donc CL = \(\dfrac{9{,}2}{\sin(57^{\circ})}\) \(\approx\) 11 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{JSH}\)) = \(\dfrac{JH}{JS}\)
\par\addvspace{.5em}
tan(55\textdegree{}) = \(\dfrac{0{,}9}{JS}\)
\par\addvspace{.5em}
 On a donc JS = \(\dfrac{0{,}9}{\tan(55^{\circ})}\) \(\approx\) 0,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{NDC}\)) = \(\dfrac{ND}{DC}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{NDC}\)) = \(\dfrac{1{,}9}{7{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{NDC}\) =ArcCos(\(\dfrac{1{,}9}{7{,}1}\)) \(\approx\)74\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{WPN}\)) = \(\dfrac{WN}{WP}\)
\par\addvspace{.5em}
d'où  tan(25\textdegree{}) =  \(\dfrac{8{,}6}{WP}\)
\par\addvspace{.5em}
On a donc WP = \(\dfrac{8{,}6}{\tan(25^{\circ})}\) \(\approx\) 18,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {3,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.553,0.760) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{ASN}\)) = \(\dfrac{AN}{AS}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{ASN}\)) =  \(\dfrac{3}{3{,}6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ASN}\) =  ArcTan(\(\dfrac{3}{3{,}6}\)) \(\approx\) 40\textdegree{}\end{document}
