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\begin{document}

\sloppy

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

 Dans le triangle ASL rectangle en A, on sait que :
\begin{itemize}
\item AL=4,6 cm
\item \(\widehat{SLA}\) =  23\textdegree{}
\end{itemize}

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

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

 Dans le triangle AWT rectangle en A, on sait que :
\begin{itemize}
\item WT=1 cm
\item \(\widehat{AWT}\) = 71\textdegree{}
\end{itemize}

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

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

 Dans le triangle SLV rectangle en S, on sait que :
\begin{itemize}
\item SL=1,7 cm
\item LV=8,8 cm
\end{itemize}

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

 Dans le triangle HJN rectangle en H, on sait que :
\begin{itemize}
\item HJ=2,7 cm
\item JN=7,5 cm
\end{itemize}

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

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

 Dans le triangle GWD rectangle en G, on sait que :
\begin{itemize}
\item GD=5,7 cm
\item \(\widehat{WDG}\) =  18\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{ALS}\)) = \(\dfrac{AS}{AL}\)
\par\addvspace{.5em}
d'où  tan(23\textdegree{}) =  \(\dfrac{AS}{4{,}6}\)
\par\addvspace{.5em}
On a donc AS =  4,6 \(\times\) tan(23\textdegree{})  \(\approx\) 2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{AWT}\)) =  \(\dfrac{AT}{WT}\)
\par\addvspace{.5em}
 d'où sin(71\textdegree{}) = \(\dfrac{AT}{1}\)

On a donc AT = 1 \(\times\) sin(71\textdegree{}) \(\approx\) 0,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{SLV}\)) = \(\dfrac{SL}{LV}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{SLV}\)) = \(\dfrac{1{,}7}{8{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{SLV}\) =ArcCos(\(\dfrac{1{,}7}{8{,}8}\)) \(\approx\)79\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{HNJ}\)) = \(\dfrac{HJ}{JN}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{HNJ}\)) =\(\dfrac{2{,}7}{7{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{HNJ}\) = ArcSin(\(\dfrac{2{,}7}{7{,}5}\)) \(\approx\) 21\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.071) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.871) {5,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.292,1.242) {18°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{GDW}\)) = \(\dfrac{GD}{WD}\)
\par\addvspace{.5em}
  d'où    cos(18\textdegree{}) =  \(\dfrac{5{,}7}{WD}\)
\par\addvspace{.5em}
 On a donc WD =  \(\dfrac{5{,}7}{\cos(18^{\circ})}\) \(\approx\) 6 cm\end{document}
