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\begin{document}

\sloppy

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

 Dans le triangle AWC rectangle en A, on sait que :
\begin{itemize}
\item AC=6,4 cm
\item WC=7,5 cm
\end{itemize}

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

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

 Dans le triangle ZBG rectangle en Z, on sait que :
\begin{itemize}
\item ZG=7,8 cm
\item \(\widehat{BGZ}\) =  23\textdegree{}
\end{itemize}

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

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

 Dans le triangle JBC rectangle en J, on sait que :
\begin{itemize}
\item JB=2,5 cm
\item JC=5,3 cm
\end{itemize}

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

 Dans le triangle KZH rectangle en K, on sait que :
\begin{itemize}
\item ZH=7,5 cm
\item \(\widehat{KZH}\) = 67\textdegree{}
\end{itemize}

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

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

 Dans le triangle RZS rectangle en R, on sait que :
\begin{itemize}
\item ZS=5,8 cm
\item \(\widehat{ZSR}\) =  21\textdegree{}
\end{itemize}

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

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

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

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.150,0.714) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{ACW}\)) = \(\dfrac{AC}{WC}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{ACW}\)) =  \(\dfrac{6{,}4}{7{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{ACW}\) = Arccos (\(\dfrac{6{,}4}{7{,}5}\)) \(\approx\) 31\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{ZGB}\)) = \(\dfrac{ZG}{BG}\)
\par\addvspace{.5em}
  d'où    cos(23\textdegree{}) =  \(\dfrac{7{,}8}{BG}\)
\par\addvspace{.5em}
 On a donc BG =  \(\dfrac{7{,}8}{\cos(23^{\circ})}\) \(\approx\) 8,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.407,2.526) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{JBC}\)) = \(\dfrac{JC}{JB}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{JBC}\)) = \(\dfrac{5{,}3}{2{,}5}\)
\par\addvspace{.5em}
 On a donc JBC = ArcTan(\(\dfrac{5{,}3}{2{,}5}\)) \(\approx\) 65\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.418,2.417) {67°};
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 Dans le triangle KZH rectangle en K, on cherche une relation entre l'angle aigu   \(\widehat{KZH}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{KZH}\)) =  \(\dfrac{KH}{ZH}\)
\par\addvspace{.5em}
 d'où sin(67\textdegree{}) = \(\dfrac{KH}{7{,}5}\)

On a donc KH = 7,5 \(\times\) sin(67\textdegree{}) \(\approx\) 6,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.926) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {5,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.294,1.114) {21°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{RSZ}\)) = \(\dfrac{RZ}{ZS}\)
\par\addvspace{.5em}
 D'où sin(21\textdegree{}) = \(\dfrac{RZ}{5{,}8}\)

On a donc RZ = 5,8 \(\times\) sin(21\textdegree{}) \(\approx\) 2,1 cm\end{document}
