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\begin{document}

\sloppy

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

 Dans le triangle MFZ rectangle en M, on sait que :
\begin{itemize}
\item FZ=2,3 cm
\item \(\widehat{FZM}\) =  41\textdegree{}
\end{itemize}

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

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

 Dans le triangle KVW rectangle en K, on sait que :
\begin{itemize}
\item KV=9,4 cm
\item \(\widehat{KVW}\) = 75\textdegree{}
\end{itemize}

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

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

 Dans le triangle JSC rectangle en J, on sait que :
\begin{itemize}
\item JC=5,6 cm
\item SC=9,3 cm
\end{itemize}

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

 Dans le triangle ZJW rectangle en Z, on sait que :
\begin{itemize}
\item ZJ=2,5 cm
\item JW=7,9 cm
\end{itemize}

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

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

 Dans le triangle BNH rectangle en B, on sait que :
\begin{itemize}
\item BH=6,1 cm
\item \(\widehat{BNH}\) = 73\textdegree{}
\end{itemize}

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

\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 (4.486,0.767) {41°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{MZF}\)) = \(\dfrac{MF}{FZ}\)
\par\addvspace{.5em}
 D'où sin(41\textdegree{}) = \(\dfrac{MF}{2{,}3}\)

On a donc MF = 2,3 \(\times\) sin(41\textdegree{}) \(\approx\) 1,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{KVW}\)) = \(\dfrac{KW}{KV}\)
\par\addvspace{.5em}
d'où tan(75\textdegree{}) =\(\dfrac{KW}{9{,}4}\)
\par\addvspace{.5em}
On a donc KW = 9,4 \(\times\) tan(75\textdegree{}) \(\approx\) 35,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{JSC}\)) =  \(\dfrac{JC}{SC}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{JSC}\)) = \(\dfrac{5{,}6}{9{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{JSC}\) =  ArcSin(\(\dfrac{5{,}6}{9{,}3}\)) \(\approx\) 37\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{ZWJ}\)) = \(\dfrac{ZJ}{JW}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{ZWJ}\)) =\(\dfrac{2{,}5}{7{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ZWJ}\) = ArcSin(\(\dfrac{2{,}5}{7{,}9}\)) \(\approx\) 18\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.447,2.146) {73°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{BNH}\)) =  \(\dfrac{BH}{NH}\)
\par\addvspace{.5em}
 d'où sin(73\textdegree{}) = \(\dfrac{6{,}1}{NH}\)
\par\addvspace{.5em}
 On a donc NH = \(\dfrac{6{,}1}{\sin(73^{\circ})}\) \(\approx\) 6,4 cm\end{document}
