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\begin{document}

\sloppy

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

 Dans le triangle RGZ rectangle en R, on sait que :
\begin{itemize}
\item GZ=8,6 cm
\item \(\widehat{GZR}\) =  31\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZRG rectangle en Z, on sait que :
\begin{itemize}
\item ZG=3,4 cm
\item \(\widehat{ZRG}\) = 80\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZWH rectangle en Z, on sait que :
\begin{itemize}
\item ZH=5,5 cm
\item \(\widehat{ZWH}\) = 54\textdegree{}
\end{itemize}

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

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

 Dans le triangle MJD rectangle en M, on sait que :
\begin{itemize}
\item MJ=2 cm
\item MD=6,4 cm
\end{itemize}

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

 Dans le triangle ZKR rectangle en Z, on sait que :
\begin{itemize}
\item ZR=5,7 cm
\item KR=7,5 cm
\end{itemize}

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

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

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

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{RZG}\)) = \(\dfrac{RZ}{GZ}\)
\par\addvspace{.5em}
  d'où    cos(31\textdegree{}) =  \(\dfrac{RZ}{8{,}6}\)

On a donc RZ = 8,6 \(\times\) cos(31\textdegree{}) \(\approx\) 7{,}4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{ZRG}\)) =  \(\dfrac{ZG}{RG}\)
\par\addvspace{.5em}
 d'où sin(80\textdegree{}) = \(\dfrac{3{,}4}{RG}\)
\par\addvspace{.5em}
 On a donc RG = \(\dfrac{3{,}4}{\sin(80^{\circ})}\) \(\approx\) 3,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{ZWH}\)) = \(\dfrac{ZH}{ZW}\)
\par\addvspace{.5em}
tan(54\textdegree{}) = \(\dfrac{5{,}5}{ZW}\)
\par\addvspace{.5em}
 On a donc ZW = \(\dfrac{5{,}5}{\tan(54^{\circ})}\) \(\approx\) 4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{MJD}\)) = \(\dfrac{MD}{MJ}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{MJD}\)) = \(\dfrac{6{,}4}{2}\)
\par\addvspace{.5em}
 On a donc MJD = ArcTan(\(\dfrac{6{,}4}{2}\)) \(\approx\) 73\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) {7,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.512,0.764) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{ZRK}\)) = \(\dfrac{ZR}{KR}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{ZRK}\)) =  \(\dfrac{5{,}7}{7{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{ZRK}\) = Arccos (\(\dfrac{5{,}7}{7{,}5}\)) \(\approx\) 41\textdegree{} \end{document}
