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\begin{document}

\sloppy

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

 Dans le triangle PAL rectangle en P, on sait que :
\begin{itemize}
\item PA=1,7 cm
\item AL=6,9 cm
\end{itemize}

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

 Dans le triangle NPF rectangle en N, on sait que :
\begin{itemize}
\item PF=3,2 cm
\item \(\widehat{PFN}\) =  22\textdegree{}
\end{itemize}

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

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

 Dans le triangle JFS rectangle en J, on sait que :
\begin{itemize}
\item JF=2,9 cm
\item JS=6,5 cm
\end{itemize}

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

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

 Dans le triangle WHN rectangle en W, on sait que :
\begin{itemize}
\item WH=9,3 cm
\item \(\widehat{WHN}\) = 79\textdegree{}
\end{itemize}

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

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

 Dans le triangle LZP rectangle en L, on sait que :
\begin{itemize}
\item LP=6,4 cm
\item \(\widehat{ZPL}\) =  11\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [PZ]. (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 (1.459,2.030) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{PAL}\)) = \(\dfrac{PA}{AL}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{PAL}\)) = \(\dfrac{1{,}7}{6{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{PAL}\) =ArcCos(\(\dfrac{1{,}7}{6{,}9}\)) \(\approx\)76\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{NFP}\)) = \(\dfrac{NP}{PF}\)
\par\addvspace{.5em}
 D'où sin(22\textdegree{}) = \(\dfrac{NP}{3{,}2}\)

On a donc NP = 3,2 \(\times\) sin(22\textdegree{}) \(\approx\) 1,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{JSF}\)) = \(\dfrac{JF}{JS}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{JSF}\)) =  \(\dfrac{2{,}9}{6{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{JSF}\) =  ArcTan(\(\dfrac{2{,}9}{6{,}5}\)) \(\approx\) 24\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{WHN}\)) = \(\dfrac{WN}{WH}\)
\par\addvspace{.5em}
d'où tan(79\textdegree{}) =\(\dfrac{WN}{9{,}3}\)
\par\addvspace{.5em}
On a donc WN = 9,3 \(\times\) tan(79\textdegree{}) \(\approx\) 47,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.190) {6,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.286,1.521) {11°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{LPZ}\)) = \(\dfrac{LP}{ZP}\)
\par\addvspace{.5em}
  d'où    cos(11\textdegree{}) =  \(\dfrac{6{,}4}{ZP}\)
\par\addvspace{.5em}
 On a donc ZP =  \(\dfrac{6{,}4}{\cos(11^{\circ})}\) \(\approx\) 6,5 cm\end{document}
