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\begin{document}

\sloppy

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

 Dans le triangle RZP rectangle en R, on sait que :
\begin{itemize}
\item RP=8,4 cm
\item \(\widehat{RZP}\) = 47\textdegree{}
\end{itemize}

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

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

 Dans le triangle VMN rectangle en V, on sait que :
\begin{itemize}
\item MN=6,3 cm
\item \(\widehat{MNV}\) =  23\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZAH rectangle en Z, on sait que :
\begin{itemize}
\item ZA=3,3 cm
\item AH=9,4 cm
\end{itemize}

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

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

 Dans le triangle NJF rectangle en N, on sait que :
\begin{itemize}
\item JF=7,1 cm
\item \(\widehat{JFN}\) =  10\textdegree{}
\end{itemize}

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

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

 Dans le triangle CVL rectangle en C, on sait que :
\begin{itemize}
\item CV=3,1 cm
\item CL=4,8 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{CVL}\). \newpage
\section*{\textcolor{red}{Correction}}

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

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

\par\addvspace{.5em}
sin(\(\widehat{RZP}\)) =  \(\dfrac{RP}{ZP}\)
\par\addvspace{.5em}
 d'où sin(47\textdegree{}) = \(\dfrac{8{,}4}{ZP}\)
\par\addvspace{.5em}
 On a donc ZP = \(\dfrac{8{,}4}{\sin(47^{\circ})}\) \(\approx\) 11,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{VNM}\)) = \(\dfrac{VN}{MN}\)
\par\addvspace{.5em}
  d'où    cos(23\textdegree{}) =  \(\dfrac{VN}{6{,}3}\)

On a donc VN = 6,3 \(\times\) cos(23\textdegree{}) \(\approx\) 5{,}8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{ZHA}\)) = \(\dfrac{ZA}{AH}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{ZHA}\)) =\(\dfrac{3{,}3}{9{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ZHA}\) = ArcSin(\(\dfrac{3{,}3}{9{,}4}\)) \(\approx\) 21\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{NFJ}\)) = \(\dfrac{NJ}{JF}\)
\par\addvspace{.5em}
 D'où sin(10\textdegree{}) = \(\dfrac{NJ}{7{,}1}\)

On a donc NJ = 7,1 \(\times\) sin(10\textdegree{}) \(\approx\) 1,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.651,2.748) {?};
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 Dans le triangle CVL rectangle en C, on cherche une relation entre l'angle aigu   \(\widehat{CVL}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{CVL}\)) = \(\dfrac{CL}{CV}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{CVL}\)) = \(\dfrac{4{,}8}{3{,}1}\)
\par\addvspace{.5em}
 On a donc CVL = ArcTan(\(\dfrac{4{,}8}{3{,}1}\)) \(\approx\) 57\textdegree{}\end{document}
