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\begin{document}

\sloppy

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

 Dans le triangle GNL rectangle en G, on sait que :
\begin{itemize}
\item NL=5,2 cm
\item \(\widehat{NLG}\) =  35\textdegree{}
\end{itemize}

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

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

 Dans le triangle PTC rectangle en P, on sait que :
\begin{itemize}
\item PC=1,8 cm
\item \(\widehat{TCP}\) =  36\textdegree{}
\end{itemize}

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

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

 Dans le triangle BFW rectangle en B, on sait que :
\begin{itemize}
\item BF=2 cm
\item BW=6,2 cm
\end{itemize}

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

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

 Dans le triangle FGT rectangle en F, on sait que :
\begin{itemize}
\item FG=2,6 cm
\item FT=3,7 cm
\end{itemize}

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

 Dans le triangle SDM rectangle en S, on sait que :
\begin{itemize}
\item SD=3,1 cm
\item \(\widehat{SDM}\) = 73\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{GLN}\)) = \(\dfrac{GL}{NL}\)
\par\addvspace{.5em}
  d'où    cos(35\textdegree{}) =  \(\dfrac{GL}{5{,}2}\)

On a donc GL = 5,2 \(\times\) cos(35\textdegree{}) \(\approx\) 4{,}3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{PCT}\)) = \(\dfrac{PT}{PC}\)
\par\addvspace{.5em}
d'où  tan(36\textdegree{}) =  \(\dfrac{PT}{1{,}8}\)
\par\addvspace{.5em}
On a donc PT =  1,8 \(\times\) tan(36\textdegree{})  \(\approx\) 1,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{BWF}\)) = \(\dfrac{BF}{BW}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{BWF}\)) =  \(\dfrac{2}{6{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{BWF}\) =  ArcTan(\(\dfrac{2}{6{,}2}\)) \(\approx\) 18\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{FGT}\)) = \(\dfrac{FT}{FG}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{FGT}\)) = \(\dfrac{3{,}7}{2{,}6}\)
\par\addvspace{.5em}
 On a donc FGT = ArcTan(\(\dfrac{3{,}7}{2{,}6}\)) \(\approx\) 55\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.447,2.146) {73°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{SDM}\)) = \(\dfrac{SD}{DM}\)
\par\addvspace{.5em}
 d'où cos(73\textdegree{}) = \(\dfrac{3{,}1}{DM}\)
\par\addvspace{.5em}
On a donc DM = \(\dfrac{3{,}1}{\cos(73^{\circ})}\)\(\approx\) 10,6 cm\end{document}
