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\begin{document}

\sloppy

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

 Dans le triangle ALR rectangle en A, on sait que :
\begin{itemize}
\item AL=2,8 cm
\item \(\widehat{ALR}\) = 52\textdegree{}
\end{itemize}

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

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

 Dans le triangle APB rectangle en A, on sait que :
\begin{itemize}
\item PB=5,8 cm
\item \(\widehat{APB}\) = 68\textdegree{}
\end{itemize}

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

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

 Dans le triangle MCG rectangle en M, on sait que :
\begin{itemize}
\item CG=5,4 cm
\item \(\widehat{CGM}\) =  40\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZTL rectangle en Z, on sait que :
\begin{itemize}
\item ZT=2,4 cm
\item ZL=6,4 cm
\end{itemize}

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

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

 Dans le triangle AGW rectangle en A, on sait que :
\begin{itemize}
\item AW=5,4 cm
\item GW=8,1 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{AGW}\). \newpage
\section*{\textcolor{red}{Correction}}

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

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

cos(\(\widehat{ALR}\)) = \(\dfrac{AL}{LR}\)
\par\addvspace{.5em}
 d'où cos(52\textdegree{}) = \(\dfrac{2{,}8}{LR}\)
\par\addvspace{.5em}
On a donc LR = \(\dfrac{2{,}8}{\cos(52^{\circ})}\)\(\approx\) 4,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{APB}\)) = \(\dfrac{AP}{PB}\)
\par\addvspace{.5em}
d'où cos(68\textdegree{}) = \(\dfrac{AP}{5{,}8}\)

On a donc AP = 5,8 \(\times\) cos(68\textdegree{}) \(\approx\) 2,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{MGC}\)) = \(\dfrac{MG}{CG}\)
\par\addvspace{.5em}
  d'où    cos(40\textdegree{}) =  \(\dfrac{MG}{5{,}4}\)

On a donc MG = 5,4 \(\times\) cos(40\textdegree{}) \(\approx\) 4{,}1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{ZLT}\)) = \(\dfrac{ZT}{ZL}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{ZLT}\)) =  \(\dfrac{2{,}4}{6{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ZLT}\) =  ArcTan(\(\dfrac{2{,}4}{6{,}4}\)) \(\approx\) 21\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.486,2.711) {?};
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 Dans le triangle AGW rectangle en A, on cherche une relation entre l'angle aigu   \(\widehat{AGW}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{AGW}\)) =  \(\dfrac{AW}{GW}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{AGW}\)) = \(\dfrac{5{,}4}{8{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{AGW}\) =  ArcSin(\(\dfrac{5{,}4}{8{,}1}\)) \(\approx\) 42\textdegree{}\end{document}
