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\begin{document}

\sloppy

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

 Dans le triangle JHN rectangle en J, on sait que :
\begin{itemize}
\item JH=6,8 cm
\item \(\widehat{JHN}\) = 74\textdegree{}
\end{itemize}

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

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

 Dans le triangle HSR rectangle en H, on sait que :
\begin{itemize}
\item HS=2,6 cm
\item SR=10 cm
\end{itemize}

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

 Dans le triangle VMP rectangle en V, on sait que :
\begin{itemize}
\item VM=9,9 cm
\item \(\widehat{MPV}\) =  34\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZJP rectangle en Z, on sait que :
\begin{itemize}
\item ZP=4 cm
\item JP=9,5 cm
\end{itemize}

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

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

 Dans le triangle WPF rectangle en W, on sait que :
\begin{itemize}
\item WF=9,6 cm
\item \(\widehat{WPF}\) = 69\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{JHN}\)) = \(\dfrac{JN}{JH}\)
\par\addvspace{.5em}
d'où tan(74\textdegree{}) =\(\dfrac{JN}{6{,}8}\)
\par\addvspace{.5em}
On a donc JN = 6,8 \(\times\) tan(74\textdegree{}) \(\approx\) 23,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{HSR}\)) = \(\dfrac{HS}{SR}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{HSR}\)) = \(\dfrac{2{,}6}{10}\)
\par\addvspace{.5em}
On a donc \(\widehat{HSR}\) =ArcCos(\(\dfrac{2{,}6}{10}\)) \(\approx\)75\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{VPM}\)) = \(\dfrac{VM}{MP}\)
\par\addvspace{.5em}
 D'où sin(34\textdegree{}) = \(\dfrac{9{,}9}{MP}\)
\par\addvspace{.5em}
On a donc MP =\(\dfrac{9{,}9}{\sin(34^{\circ})}\) \(\approx\) 17,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{ZPJ}\)) = \(\dfrac{ZP}{JP}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{ZPJ}\)) =  \(\dfrac{4}{9{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{ZPJ}\) = Arccos (\(\dfrac{4}{9{,}5}\)) \(\approx\) 65\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.428,2.324) {69°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{WPF}\)) = \(\dfrac{WF}{WP}\)
\par\addvspace{.5em}
tan(69\textdegree{}) = \(\dfrac{9{,}6}{WP}\)
\par\addvspace{.5em}
 On a donc WP = \(\dfrac{9{,}6}{\tan(69^{\circ})}\) \(\approx\) 3,7 cm\end{document}
