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\begin{document}

\sloppy

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

 Dans le triangle WSC rectangle en W, on sait que :
\begin{itemize}
\item WC=5,7 cm
\item \(\widehat{WSC}\) = 52\textdegree{}
\end{itemize}

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

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

 Dans le triangle BSF rectangle en B, on sait que :
\begin{itemize}
\item BS=2,3 cm
\item SF=7,3 cm
\end{itemize}

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

 Dans le triangle KGF rectangle en K, on sait que :
\begin{itemize}
\item KG=6,2 cm
\item \(\widehat{KGF}\) = 80\textdegree{}
\end{itemize}

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

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

 Dans le triangle JHN rectangle en J, on sait que :
\begin{itemize}
\item HN=5,4 cm
\item \(\widehat{HNJ}\) =  45\textdegree{}
\end{itemize}

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

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

 Dans le triangle JHA rectangle en J, on sait que :
\begin{itemize}
\item JH=2,5 cm
\item HA=7,6 cm
\end{itemize}

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

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

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

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.000,2.734) {52°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{WSC}\)) =  \(\dfrac{WC}{SC}\)
\par\addvspace{.5em}
 d'où sin(52\textdegree{}) = \(\dfrac{5{,}7}{SC}\)
\par\addvspace{.5em}
 On a donc SC = \(\dfrac{5{,}7}{\sin(52^{\circ})}\) \(\approx\) 7,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{BSF}\)) = \(\dfrac{BS}{SF}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{BSF}\)) = \(\dfrac{2{,}3}{7{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{BSF}\) =ArcCos(\(\dfrac{2{,}3}{7{,}3}\)) \(\approx\)72\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{KGF}\)) = \(\dfrac{KF}{KG}\)
\par\addvspace{.5em}
d'où tan(80\textdegree{}) =\(\dfrac{KF}{6{,}2}\)
\par\addvspace{.5em}
On a donc KF = 6,2 \(\times\) tan(80\textdegree{}) \(\approx\) 35,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \draw[black] (2.380,0.880) -- (2.386,0.788) -- (2.382,0.694) -- (2.380,0.600);
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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{JNH}\)) = \(\dfrac{JH}{HN}\)
\par\addvspace{.5em}
 D'où sin(45\textdegree{}) = \(\dfrac{JH}{5{,}4}\)

On a donc JH = 5,4 \(\times\) sin(45\textdegree{}) \(\approx\) 3,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.293,1.191) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{JAH}\)) = \(\dfrac{JH}{HA}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{JAH}\)) =\(\dfrac{2{,}5}{7{,}6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{JAH}\) = ArcSin(\(\dfrac{2{,}5}{7{,}6}\)) \(\approx\) 19\textdegree{}\end{document}
