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\begin{document}

\sloppy

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

 Dans le triangle JDP rectangle en J, on sait que :
\begin{itemize}
\item JP=1,5 cm
\item \(\widehat{JDP}\) = 71\textdegree{}
\end{itemize}

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

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

 Dans le triangle FAM rectangle en F, on sait que :
\begin{itemize}
\item FA=4,9 cm
\item \(\widehat{FAM}\) = 50\textdegree{}
\end{itemize}

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

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

 Dans le triangle BMJ rectangle en B, on sait que :
\begin{itemize}
\item BM=2,3 cm
\item MJ=7,6 cm
\end{itemize}

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

 Dans le triangle TSB rectangle en T, on sait que :
\begin{itemize}
\item TB=3,5 cm
\item \(\widehat{TSB}\) = 48\textdegree{}
\end{itemize}

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

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

 Dans le triangle RFG rectangle en R, on sait que :
\begin{itemize}
\item RF=2,8 cm
\item RG=5,3 cm
\end{itemize}

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

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

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

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{JDP}\)) = \(\dfrac{JP}{JD}\)
\par\addvspace{.5em}
tan(71\textdegree{}) = \(\dfrac{1{,}5}{JD}\)
\par\addvspace{.5em}
 On a donc JD = \(\dfrac{1{,}5}{\tan(71^{\circ})}\) \(\approx\) 0,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{FAM}\)) = \(\dfrac{FM}{FA}\)
\par\addvspace{.5em}
d'où tan(50\textdegree{}) =\(\dfrac{FM}{4{,}9}\)
\par\addvspace{.5em}
On a donc FM = 4,9 \(\times\) tan(50\textdegree{}) \(\approx\) 5,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{BMJ}\)) = \(\dfrac{BM}{MJ}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{BMJ}\)) = \(\dfrac{2{,}3}{7{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{BMJ}\) =ArcCos(\(\dfrac{2{,}3}{7{,}6}\)) \(\approx\)72\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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\end{tikzpicture}\end{center}


 Dans le triangle TSB rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TSB}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{TSB}\)) =  \(\dfrac{TB}{SB}\)
\par\addvspace{.5em}
 d'où sin(48\textdegree{}) = \(\dfrac{3{,}5}{SB}\)
\par\addvspace{.5em}
 On a donc SB = \(\dfrac{3{,}5}{\sin(48^{\circ})}\) \(\approx\) 4,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.372) {5,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.303,0.799) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{RGF}\)) = \(\dfrac{RF}{RG}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{RGF}\)) =  \(\dfrac{2{,}8}{5{,}3}\)
\par\addvspace{.5em}
 On a donc \(\widehat{RGF}\) =  ArcTan(\(\dfrac{2{,}8}{5{,}3}\)) \(\approx\) 28\textdegree{}\end{document}
