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\begin{document}

\sloppy

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

 Dans le triangle GVA rectangle en G, on sait que :
\begin{itemize}
\item GA=4,9 cm
\item \(\widehat{VAG}\) =  35\textdegree{}
\end{itemize}

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

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

 Dans le triangle ARZ rectangle en A, on sait que :
\begin{itemize}
\item AR=4,3 cm
\item \(\widehat{ARZ}\) = 59\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZTR rectangle en Z, on sait que :
\begin{itemize}
\item ZT=1,2 cm
\item ZR=4,4 cm
\end{itemize}

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

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

 Dans le triangle CLR rectangle en C, on sait que :
\begin{itemize}
\item CL=1,8 cm
\item CR=6,2 cm
\end{itemize}

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

 Dans le triangle RGZ rectangle en R, on sait que :
\begin{itemize}
\item GZ=4 cm
\item \(\widehat{RGZ}\) = 72\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{GAV}\)) = \(\dfrac{GA}{VA}\)
\par\addvspace{.5em}
  d'où    cos(35\textdegree{}) =  \(\dfrac{4{,}9}{VA}\)
\par\addvspace{.5em}
 On a donc VA =  \(\dfrac{4{,}9}{\cos(35^{\circ})}\) \(\approx\) 6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{ARZ}\)) = \(\dfrac{AZ}{AR}\)
\par\addvspace{.5em}
d'où tan(59\textdegree{}) =\(\dfrac{AZ}{4{,}3}\)
\par\addvspace{.5em}
On a donc AZ = 4,3 \(\times\) tan(59\textdegree{}) \(\approx\) 7,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{ZRT}\)) = \(\dfrac{ZT}{ZR}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{ZRT}\)) =  \(\dfrac{1{,}2}{4{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ZRT}\) =  ArcTan(\(\dfrac{1{,}2}{4{,}4}\)) \(\approx\) 15\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{CLR}\)) = \(\dfrac{CR}{CL}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{CLR}\)) = \(\dfrac{6{,}2}{1{,}8}\)
\par\addvspace{.5em}
 On a donc CLR = ArcTan(\(\dfrac{6{,}2}{1{,}8}\)) \(\approx\) 74\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {?};
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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.442,2.190) {72°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{RGZ}\)) = \(\dfrac{RG}{GZ}\)
\par\addvspace{.5em}
d'où cos(72\textdegree{}) = \(\dfrac{RG}{4}\)

On a donc RG = 4 \(\times\) cos(72\textdegree{}) \(\approx\) 1,2 cm\end{document}
