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\begin{document}

\sloppy

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

 Dans le triangle PZA rectangle en P, on sait que :
\begin{itemize}
\item PZ=1,4 cm
\item ZA=7,6 cm
\end{itemize}

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

 Dans le triangle MVP rectangle en M, on sait que :
\begin{itemize}
\item VP=7,1 cm
\item \(\widehat{VPM}\) =  30\textdegree{}
\end{itemize}

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

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

 Dans le triangle HLV rectangle en H, on sait que :
\begin{itemize}
\item LV=0,6 cm
\item \(\widehat{LVH}\) =  24\textdegree{}
\end{itemize}

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

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

 Dans le triangle LJR rectangle en L, on sait que :
\begin{itemize}
\item LJ=5,3 cm
\item \(\widehat{JRL}\) =  36\textdegree{}
\end{itemize}

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

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

 Dans le triangle CNF rectangle en C, on sait que :
\begin{itemize}
\item CF=4,8 cm
\item NF=8,9 cm
\end{itemize}

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

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

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

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

cos(\(\widehat{PZA}\)) = \(\dfrac{PZ}{ZA}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{PZA}\)) = \(\dfrac{1{,}4}{7{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{PZA}\) =ArcCos(\(\dfrac{1{,}4}{7{,}6}\)) \(\approx\)79\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{MPV}\)) = \(\dfrac{MV}{VP}\)
\par\addvspace{.5em}
 D'où sin(30\textdegree{}) = \(\dfrac{MV}{7{,}1}\)

On a donc MV = 7,1 \(\times\) sin(30\textdegree{}) \(\approx\) 3,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{HVL}\)) = \(\dfrac{HV}{LV}\)
\par\addvspace{.5em}
  d'où    cos(24\textdegree{}) =  \(\dfrac{HV}{0{,}6}\)

On a donc HV = 0,6 \(\times\) cos(24\textdegree{}) \(\approx\) 0{,}5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{LRJ}\)) = \(\dfrac{LJ}{JR}\)
\par\addvspace{.5em}
 D'où sin(36\textdegree{}) = \(\dfrac{5{,}3}{JR}\)
\par\addvspace{.5em}
On a donc JR =\(\dfrac{5{,}3}{\sin(36^{\circ})}\) \(\approx\) 9 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) {8,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.812,0.853) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{CFN}\)) = \(\dfrac{CF}{NF}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{CFN}\)) =  \(\dfrac{4{,}8}{8{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{CFN}\) = Arccos (\(\dfrac{4{,}8}{8{,}9}\)) \(\approx\) 57\textdegree{} \end{document}
