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\begin{document}

\sloppy

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

 Dans le triangle ZGS rectangle en Z, on sait que :
\begin{itemize}
\item ZS=4,8 cm
\item \(\widehat{ZGS}\) = 69\textdegree{}
\end{itemize}

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

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

 Dans le triangle CHF rectangle en C, on sait que :
\begin{itemize}
\item CH=5,2 cm
\item \(\widehat{HFC}\) =  29\textdegree{}
\end{itemize}

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

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

 Dans le triangle TJA rectangle en T, on sait que :
\begin{itemize}
\item TJ=7,5 cm
\item \(\widehat{TJA}\) = 60\textdegree{}
\end{itemize}

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

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

 Dans le triangle JAH rectangle en J, on sait que :
\begin{itemize}
\item JA=1,3 cm
\item JH=5,5 cm
\end{itemize}

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

 Dans le triangle MSB rectangle en M, on sait que :
\begin{itemize}
\item MS=3,1 cm
\item SB=8,8 cm
\end{itemize}

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

\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 (1.428,2.324) {69°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{ZGS}\)) = \(\dfrac{ZS}{ZG}\)
\par\addvspace{.5em}
tan(69\textdegree{}) = \(\dfrac{4{,}8}{ZG}\)
\par\addvspace{.5em}
 On a donc ZG = \(\dfrac{4{,}8}{\tan(69^{\circ})}\) \(\approx\) 1,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{CFH}\)) = \(\dfrac{CH}{CF}\)
\par\addvspace{.5em}
d'où  tan(29\textdegree{}) =  \(\dfrac{5{,}2}{CF}\)
\par\addvspace{.5em}
On a donc CF = \(\dfrac{5{,}2}{\tan(29^{\circ})}\) \(\approx\) 9,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.408,2.756) {60°};
\end{tikzpicture}\end{center}


 Dans le triangle TJA rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TJA}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{TJA}\)) = \(\dfrac{TJ}{JA}\)
\par\addvspace{.5em}
 d'où cos(60\textdegree{}) = \(\dfrac{7{,}5}{JA}\)
\par\addvspace{.5em}
On a donc JA = \(\dfrac{7{,}5}{\cos(60^{\circ})}\)\(\approx\) 15 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{JAH}\)) = \(\dfrac{JH}{JA}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{JAH}\)) = \(\dfrac{5{,}5}{1{,}3}\)
\par\addvspace{.5em}
 On a donc JAH = ArcTan(\(\dfrac{5{,}5}{1{,}3}\)) \(\approx\) 77\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) {3,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.294,1.131) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{MBS}\)) = \(\dfrac{MS}{SB}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{MBS}\)) =\(\dfrac{3{,}1}{8{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{MBS}\) = ArcSin(\(\dfrac{3{,}1}{8{,}8}\)) \(\approx\) 21\textdegree{}\end{document}
