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\begin{document}

\sloppy

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

 Dans le triangle WLN rectangle en W, on sait que :
\begin{itemize}
\item WN=2,5 cm
\item \(\widehat{LNW}\) =  38\textdegree{}
\end{itemize}

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

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

 Dans le triangle BAF rectangle en B, on sait que :
\begin{itemize}
\item BA=1,8 cm
\item BF=5,9 cm
\end{itemize}

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

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

 Dans le triangle VHA rectangle en V, on sait que :
\begin{itemize}
\item HA=5,7 cm
\item \(\widehat{HAV}\) =  13\textdegree{}
\end{itemize}

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

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

 Dans le triangle SPZ rectangle en S, on sait que :
\begin{itemize}
\item PZ=4,9 cm
\item \(\widehat{SPZ}\) = 66\textdegree{}
\end{itemize}

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

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

 Dans le triangle JKC rectangle en J, on sait que :
\begin{itemize}
\item JK=2,4 cm
\item JC=5 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{JKC}\). \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 (4.662,0.750) {38°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{WNL}\)) = \(\dfrac{WN}{LN}\)
\par\addvspace{.5em}
  d'où    cos(38\textdegree{}) =  \(\dfrac{2{,}5}{LN}\)
\par\addvspace{.5em}
 On a donc LN =  \(\dfrac{2{,}5}{\cos(38^{\circ})}\) \(\approx\) 3,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{BFA}\)) = \(\dfrac{BA}{BF}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{BFA}\)) =  \(\dfrac{1{,}8}{5{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{BFA}\) =  ArcTan(\(\dfrac{1{,}8}{5{,}9}\)) \(\approx\) 17\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{VAH}\)) = \(\dfrac{VA}{HA}\)
\par\addvspace{.5em}
  d'où    cos(13\textdegree{}) =  \(\dfrac{VA}{5{,}7}\)

On a donc VA = 5,7 \(\times\) cos(13\textdegree{}) \(\approx\) 5{,}6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{SPZ}\)) = \(\dfrac{SP}{PZ}\)
\par\addvspace{.5em}
d'où cos(66\textdegree{}) = \(\dfrac{SP}{4{,}9}\)

On a donc SP = 4,9 \(\times\) cos(66\textdegree{}) \(\approx\) 2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.691) {C};
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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.491) {5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.405,2.545) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{JKC}\)) = \(\dfrac{JC}{JK}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{JKC}\)) = \(\dfrac{5}{2{,}4}\)
\par\addvspace{.5em}
 On a donc JKC = ArcTan(\(\dfrac{5}{2{,}4}\)) \(\approx\) 64\textdegree{}\end{document}
