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\begin{document}

\sloppy

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

 Dans le triangle NFL rectangle en N, on sait que :
\begin{itemize}
\item NL=2,3 cm
\item \(\widehat{NFL}\) = 69\textdegree{}
\end{itemize}

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

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

 Dans le triangle WPJ rectangle en W, on sait que :
\begin{itemize}
\item WJ=0,5 cm
\item \(\widehat{PJW}\) =  34\textdegree{}
\end{itemize}

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

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

 Dans le triangle PVN rectangle en P, on sait que :
\begin{itemize}
\item PV=3 cm
\item PN=5,2 cm
\end{itemize}

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

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

 Dans le triangle JSV rectangle en J, on sait que :
\begin{itemize}
\item JV=5 cm
\item SV=7,9 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{JSV}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle MFW rectangle en M, on sait que :
\begin{itemize}
\item MF=8,5 cm
\item \(\widehat{FWM}\) =  29\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [MW]. (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 (1.428,2.324) {69°};
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 Dans le triangle NFL rectangle en N, on cherche une relation entre l'angle aigu   \(\widehat{NFL}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{NFL}\)) = \(\dfrac{NL}{NF}\)
\par\addvspace{.5em}
tan(69\textdegree{}) = \(\dfrac{2{,}3}{NF}\)
\par\addvspace{.5em}
 On a donc NF = \(\dfrac{2{,}3}{\tan(69^{\circ})}\) \(\approx\) 0,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{WJP}\)) = \(\dfrac{WJ}{PJ}\)
\par\addvspace{.5em}
  d'où    cos(34\textdegree{}) =  \(\dfrac{0{,}5}{PJ}\)
\par\addvspace{.5em}
 On a donc PJ =  \(\dfrac{0{,}5}{\cos(34^{\circ})}\) \(\approx\) 0,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{PNV}\)) = \(\dfrac{PV}{PN}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{PNV}\)) =  \(\dfrac{3}{5{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{PNV}\) =  ArcTan(\(\dfrac{3}{5{,}2}\)) \(\approx\) 30\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{JSV}\)) =  \(\dfrac{JV}{SV}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{JSV}\)) = \(\dfrac{5}{7{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{JSV}\) =  ArcSin(\(\dfrac{5}{7{,}9}\)) \(\approx\) 39\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.309) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.305,0.742) {29°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{MWF}\)) = \(\dfrac{MF}{MW}\)
\par\addvspace{.5em}
d'où  tan(29\textdegree{}) =  \(\dfrac{8{,}5}{MW}\)
\par\addvspace{.5em}
On a donc MW = \(\dfrac{8{,}5}{\tan(29^{\circ})}\) \(\approx\) 15,3 cm\end{document}
