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\begin{document}

\sloppy

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

 Dans le triangle TCZ rectangle en T, on sait que :
\begin{itemize}
\item TZ=9,4 cm
\item \(\widehat{CZT}\) =  21\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZLN rectangle en Z, on sait que :
\begin{itemize}
\item ZL=0,4 cm
\item \(\widehat{LNZ}\) =  25\textdegree{}
\end{itemize}

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

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

 Dans le triangle DSJ rectangle en D, on sait que :
\begin{itemize}
\item DS=1,9 cm
\item SJ=7,5 cm
\end{itemize}

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

 Dans le triangle RAC rectangle en R, on sait que :
\begin{itemize}
\item RA=1,2 cm
\item AC=8,7 cm
\end{itemize}

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

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

 Dans le triangle HGA rectangle en H, on sait que :
\begin{itemize}
\item HG=5,4 cm
\item \(\widehat{HGA}\) = 65\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{TZC}\)) = \(\dfrac{TC}{TZ}\)
\par\addvspace{.5em}
d'où  tan(21\textdegree{}) =  \(\dfrac{TC}{9{,}4}\)
\par\addvspace{.5em}
On a donc TC =  9,4 \(\times\) tan(21\textdegree{})  \(\approx\) 3,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{ZNL}\)) = \(\dfrac{ZL}{ZN}\)
\par\addvspace{.5em}
d'où  tan(25\textdegree{}) =  \(\dfrac{0{,}4}{ZN}\)
\par\addvspace{.5em}
On a donc ZN = \(\dfrac{0{,}4}{\tan(25^{\circ})}\) \(\approx\) 0,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{DSJ}\)) = \(\dfrac{DS}{SJ}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{DSJ}\)) = \(\dfrac{1{,}9}{7{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{DSJ}\) =ArcCos(\(\dfrac{1{,}9}{7{,}5}\)) \(\approx\)75\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{RCA}\)) = \(\dfrac{RA}{AC}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{RCA}\)) =\(\dfrac{1{,}2}{8{,}7}\)
\par\addvspace{.5em}
 On a donc \(\widehat{RCA}\) = ArcSin(\(\dfrac{1{,}2}{8{,}7}\)) \(\approx\) 8\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) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.408,2.514) {65°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{HGA}\)) = \(\dfrac{HG}{GA}\)
\par\addvspace{.5em}
 d'où cos(65\textdegree{}) = \(\dfrac{5{,}4}{GA}\)
\par\addvspace{.5em}
On a donc GA = \(\dfrac{5{,}4}{\cos(65^{\circ})}\)\(\approx\) 12,8 cm\end{document}
