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\begin{document}

\sloppy

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

 Dans le triangle LST rectangle en L, on sait que :
\begin{itemize}
\item LT=5,8 cm
\item ST=9,3 cm
\end{itemize}

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

 Dans le triangle SFV rectangle en S, on sait que :
\begin{itemize}
\item FV=8,5 cm
\item \(\widehat{FVS}\) =  18\textdegree{}
\end{itemize}

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

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

 Dans le triangle WBD rectangle en W, on sait que :
\begin{itemize}
\item WB=1,6 cm
\item BD=8,5 cm
\end{itemize}

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

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

 Dans le triangle HAP rectangle en H, on sait que :
\begin{itemize}
\item AP=3,4 cm
\item \(\widehat{HAP}\) = 73\textdegree{}
\end{itemize}

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

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

 Dans le triangle MJW rectangle en M, on sait que :
\begin{itemize}
\item MJ=1,3 cm
\item \(\widehat{MJW}\) = 62\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{LST}\)) =  \(\dfrac{LT}{ST}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{LST}\)) = \(\dfrac{5{,}8}{9{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{LST}\) =  ArcSin(\(\dfrac{5{,}8}{9{,}3}\)) \(\approx\) 39\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{SVF}\)) = \(\dfrac{SF}{FV}\)
\par\addvspace{.5em}
 D'où sin(18\textdegree{}) = \(\dfrac{SF}{8{,}5}\)

On a donc SF = 8,5 \(\times\) sin(18\textdegree{}) \(\approx\) 2,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{WDB}\)) = \(\dfrac{WB}{BD}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{WDB}\)) =\(\dfrac{1{,}6}{8{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{WDB}\) = ArcSin(\(\dfrac{1{,}6}{8{,}5}\)) \(\approx\) 11\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{HAP}\)) =  \(\dfrac{HP}{AP}\)
\par\addvspace{.5em}
 d'où sin(73\textdegree{}) = \(\dfrac{HP}{3{,}4}\)

On a donc HP = 3,4 \(\times\) sin(73\textdegree{}) \(\approx\) 3,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \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 (1.393,2.665) {62°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{MJW}\)) = \(\dfrac{MJ}{JW}\)
\par\addvspace{.5em}
 d'où cos(62\textdegree{}) = \(\dfrac{1{,}3}{JW}\)
\par\addvspace{.5em}
On a donc JW = \(\dfrac{1{,}3}{\cos(62^{\circ})}\)\(\approx\) 2,8 cm\end{document}
