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\begin{document}

\sloppy

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

 Dans le triangle APT rectangle en A, on sait que :
\begin{itemize}
\item AP=1,1 cm
\item AT=4,9 cm
\end{itemize}

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

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

 Dans le triangle GFL rectangle en G, on sait que :
\begin{itemize}
\item GL=7,7 cm
\item \(\widehat{GFL}\) = 79\textdegree{}
\end{itemize}

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

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

 Dans le triangle JDV rectangle en J, on sait que :
\begin{itemize}
\item JD=7 cm
\item \(\widehat{JDV}\) = 59\textdegree{}
\end{itemize}

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

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

 Dans le triangle MJA rectangle en M, on sait que :
\begin{itemize}
\item MA=4,8 cm
\item JA=7,6 cm
\end{itemize}

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

 Dans le triangle LKV rectangle en L, on sait que :
\begin{itemize}
\item KV=4,8 cm
\item \(\widehat{KVL}\) =  10\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{ATP}\)) = \(\dfrac{AP}{AT}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{ATP}\)) =  \(\dfrac{1{,}1}{4{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ATP}\) =  ArcTan(\(\dfrac{1{,}1}{4{,}9}\)) \(\approx\) 13\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{GFL}\)) =  \(\dfrac{GL}{FL}\)
\par\addvspace{.5em}
 d'où sin(79\textdegree{}) = \(\dfrac{7{,}7}{FL}\)
\par\addvspace{.5em}
 On a donc FL = \(\dfrac{7{,}7}{\sin(79^{\circ})}\) \(\approx\) 7,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{JDV}\)) = \(\dfrac{JV}{JD}\)
\par\addvspace{.5em}
d'où tan(59\textdegree{}) =\(\dfrac{JV}{7}\)
\par\addvspace{.5em}
On a donc JV = 7 \(\times\) tan(59\textdegree{}) \(\approx\) 11,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{MJA}\)) =  \(\dfrac{MA}{JA}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{MJA}\)) = \(\dfrac{4{,}8}{7{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{MJA}\) =  ArcSin(\(\dfrac{4{,}8}{7{,}6}\)) \(\approx\) 39\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.286,1.559) {10°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{LVK}\)) = \(\dfrac{LK}{KV}\)
\par\addvspace{.5em}
 D'où sin(10\textdegree{}) = \(\dfrac{LK}{4{,}8}\)

On a donc LK = 4,8 \(\times\) sin(10\textdegree{}) \(\approx\) 0,8 cm\end{document}
