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\begin{document}

\sloppy

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

 Dans le triangle LMD rectangle en L, on sait que :
\begin{itemize}
\item LM=5,8 cm
\item \(\widehat{LMD}\) = 69\textdegree{}
\end{itemize}

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

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

 Dans le triangle FLT rectangle en F, on sait que :
\begin{itemize}
\item LT=8,4 cm
\item \(\widehat{FLT}\) = 59\textdegree{}
\end{itemize}

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

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

 Dans le triangle BDL rectangle en B, on sait que :
\begin{itemize}
\item BD=3,3 cm
\item BL=4,8 cm
\end{itemize}

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

 Dans le triangle BJF rectangle en B, on sait que :
\begin{itemize}
\item JF=0,6 cm
\item \(\widehat{JFB}\) =  17\textdegree{}
\end{itemize}

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

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

 Dans le triangle GBK rectangle en G, on sait que :
\begin{itemize}
\item GB=2,9 cm
\item GK=6 cm
\end{itemize}

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

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

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

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

cos(\(\widehat{LMD}\)) = \(\dfrac{LM}{MD}\)
\par\addvspace{.5em}
 d'où cos(69\textdegree{}) = \(\dfrac{5{,}8}{MD}\)
\par\addvspace{.5em}
On a donc MD = \(\dfrac{5{,}8}{\cos(69^{\circ})}\)\(\approx\) 16,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{FLT}\)) = \(\dfrac{FL}{LT}\)
\par\addvspace{.5em}
d'où cos(59\textdegree{}) = \(\dfrac{FL}{8{,}4}\)

On a donc FL = 8,4 \(\times\) cos(59\textdegree{}) \(\approx\) 4,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{BDL}\)) = \(\dfrac{BL}{BD}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{BDL}\)) = \(\dfrac{4{,}8}{3{,}3}\)
\par\addvspace{.5em}
 On a donc BDL = ArcTan(\(\dfrac{4{,}8}{3{,}3}\)) \(\approx\) 55\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{BFJ}\)) = \(\dfrac{BF}{JF}\)
\par\addvspace{.5em}
  d'où    cos(17\textdegree{}) =  \(\dfrac{BF}{0{,}6}\)

On a donc BF = 0,6 \(\times\) cos(17\textdegree{}) \(\approx\) 0{,}6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.300,0.898) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{GKB}\)) = \(\dfrac{GB}{GK}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{GKB}\)) =  \(\dfrac{2{,}9}{6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{GKB}\) =  ArcTan(\(\dfrac{2{,}9}{6}\)) \(\approx\) 26\textdegree{}\end{document}
