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\begin{document}

\sloppy

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

 Dans le triangle WHJ rectangle en W, on sait que :
\begin{itemize}
\item WH=1,7 cm
\item HJ=7,3 cm
\end{itemize}

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

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

 Dans le triangle HNW rectangle en H, on sait que :
\begin{itemize}
\item HW=5,2 cm
\item NW=8,4 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{HNW}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle TGD rectangle en T, on sait que :
\begin{itemize}
\item GD=8,4 cm
\item \(\widehat{TGD}\) = 70\textdegree{}
\end{itemize}

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

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

 Dans le triangle NWS rectangle en N, on sait que :
\begin{itemize}
\item WS=6,4 cm
\item \(\widehat{WSN}\) =  38\textdegree{}
\end{itemize}

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

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

 Dans le triangle SVA rectangle en S, on sait que :
\begin{itemize}
\item SV=5,4 cm
\item \(\widehat{VAS}\) =  36\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{WJH}\)) = \(\dfrac{WH}{HJ}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{WJH}\)) =\(\dfrac{1{,}7}{7{,}3}\)
\par\addvspace{.5em}
 On a donc \(\widehat{WJH}\) = ArcSin(\(\dfrac{1{,}7}{7{,}3}\)) \(\approx\) 13\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{HNW}\)) =  \(\dfrac{HW}{NW}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{HNW}\)) = \(\dfrac{5{,}2}{8{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{HNW}\) =  ArcSin(\(\dfrac{5{,}2}{8{,}4}\)) \(\approx\) 38\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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 Dans le triangle TGD rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TGD}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{TGD}\)) =  \(\dfrac{TD}{GD}\)
\par\addvspace{.5em}
 d'où sin(70\textdegree{}) = \(\dfrac{TD}{8{,}4}\)

On a donc TD = 8,4 \(\times\) sin(70\textdegree{}) \(\approx\) 7,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.662,0.750) {38°};
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 Dans le triangle NWS rectangle en N, on cherche une relation entre l'angle aigu   \(\widehat{NSW}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{NSW}\)) = \(\dfrac{NW}{WS}\)
\par\addvspace{.5em}
 D'où sin(38\textdegree{}) = \(\dfrac{NW}{6{,}4}\)

On a donc NW = 6,4 \(\times\) sin(38\textdegree{}) \(\approx\) 3,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.793,0.739) {36°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{SAV}\)) = \(\dfrac{SV}{VA}\)
\par\addvspace{.5em}
 D'où sin(36\textdegree{}) = \(\dfrac{5{,}4}{VA}\)
\par\addvspace{.5em}
On a donc VA =\(\dfrac{5{,}4}{\sin(36^{\circ})}\) \(\approx\) 9,2 cm\end{document}
