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\begin{document}

\sloppy

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

 Dans le triangle LGD rectangle en L, on sait que :
\begin{itemize}
\item LG=9,1 cm
\item \(\widehat{GDL}\) =  27\textdegree{}
\end{itemize}

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

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

 Dans le triangle GBT rectangle en G, on sait que :
\begin{itemize}
\item GT=4,4 cm
\item BT=6,8 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{GBT}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle SBM rectangle en S, on sait que :
\begin{itemize}
\item BM=1,3 cm
\item \(\widehat{SBM}\) = 74\textdegree{}
\end{itemize}

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

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

 Dans le triangle WLN rectangle en W, on sait que :
\begin{itemize}
\item WL=1,6 cm
\item LN=8,4 cm
\end{itemize}

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

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

 Dans le triangle RNC rectangle en R, on sait que :
\begin{itemize}
\item RC=6,1 cm
\item \(\widehat{RNC}\) = 70\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{LDG}\)) = \(\dfrac{LG}{LD}\)
\par\addvspace{.5em}
d'où  tan(27\textdegree{}) =  \(\dfrac{9{,}1}{LD}\)
\par\addvspace{.5em}
On a donc LD = \(\dfrac{9{,}1}{\tan(27^{\circ})}\) \(\approx\) 17,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{GBT}\)) =  \(\dfrac{GT}{BT}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{GBT}\)) = \(\dfrac{4{,}4}{6{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{GBT}\) =  ArcSin(\(\dfrac{4{,}4}{6{,}8}\)) \(\approx\) 40\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{SBM}\)) = \(\dfrac{SB}{BM}\)
\par\addvspace{.5em}
d'où cos(74\textdegree{}) = \(\dfrac{SB}{1{,}3}\)

On a donc SB = 1,3 \(\times\) cos(74\textdegree{}) \(\approx\) 0,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{WNL}\)) = \(\dfrac{WL}{LN}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{WNL}\)) =\(\dfrac{1{,}6}{8{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{WNL}\) = ArcSin(\(\dfrac{1{,}6}{8{,}4}\)) \(\approx\) 11\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.432,2.279) {70°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{RNC}\)) =  \(\dfrac{RC}{NC}\)
\par\addvspace{.5em}
 d'où sin(70\textdegree{}) = \(\dfrac{6{,}1}{NC}\)
\par\addvspace{.5em}
 On a donc NC = \(\dfrac{6{,}1}{\sin(70^{\circ})}\) \(\approx\) 6,5 cm\end{document}
