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\begin{document}

\sloppy

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

 Dans le triangle JBR rectangle en J, on sait que :
\begin{itemize}
\item JB=2,6 cm
\item JR=5,3 cm
\end{itemize}

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

 Dans le triangle RWL rectangle en R, on sait que :
\begin{itemize}
\item WL=3,3 cm
\item \(\widehat{RWL}\) = 53\textdegree{}
\end{itemize}

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

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

 Dans le triangle FPZ rectangle en F, on sait que :
\begin{itemize}
\item FP=1,6 cm
\item FZ=4,8 cm
\end{itemize}

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

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

 Dans le triangle MNA rectangle en M, on sait que :
\begin{itemize}
\item MA=2,2 cm
\item \(\widehat{NAM}\) =  19\textdegree{}
\end{itemize}

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

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

 Dans le triangle AWL rectangle en A, on sait que :
\begin{itemize}
\item AW=9,8 cm
\item \(\widehat{AWL}\) = 65\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{JBR}\)) = \(\dfrac{JR}{JB}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{JBR}\)) = \(\dfrac{5{,}3}{2{,}6}\)
\par\addvspace{.5em}
 On a donc JBR = ArcTan(\(\dfrac{5{,}3}{2{,}6}\)) \(\approx\) 64\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{RWL}\)) = \(\dfrac{RW}{WL}\)
\par\addvspace{.5em}
d'où cos(53\textdegree{}) = \(\dfrac{RW}{3{,}3}\)

On a donc RW = 3,3 \(\times\) cos(53\textdegree{}) \(\approx\) 2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{FZP}\)) = \(\dfrac{FP}{FZ}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{FZP}\)) =  \(\dfrac{1{,}6}{4{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{FZP}\) =  ArcTan(\(\dfrac{1{,}6}{4{,}8}\)) \(\approx\) 18\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{MAN}\)) = \(\dfrac{MA}{NA}\)
\par\addvspace{.5em}
  d'où    cos(19\textdegree{}) =  \(\dfrac{2{,}2}{NA}\)
\par\addvspace{.5em}
 On a donc NA =  \(\dfrac{2{,}2}{\cos(19^{\circ})}\) \(\approx\) 2,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.524) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.408,2.514) {65°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{AWL}\)) = \(\dfrac{AL}{AW}\)
\par\addvspace{.5em}
d'où tan(65\textdegree{}) =\(\dfrac{AL}{9{,}8}\)
\par\addvspace{.5em}
On a donc AL = 9,8 \(\times\) tan(65\textdegree{}) \(\approx\) 21 cm\end{document}
