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\begin{document}

\sloppy

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

 Dans le triangle ABJ rectangle en A, on sait que :
\begin{itemize}
\item AB=3,3 cm
\item AJ=4,4 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{ABJ}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle ZTR rectangle en Z, on sait que :
\begin{itemize}
\item TR=8,7 cm
\item \(\widehat{ZTR}\) = 64\textdegree{}
\end{itemize}

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

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

 Dans le triangle MDS rectangle en M, on sait que :
\begin{itemize}
\item DS=1,1 cm
\item \(\widehat{DSM}\) =  16\textdegree{}
\end{itemize}

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

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

 Dans le triangle ATF rectangle en A, on sait que :
\begin{itemize}
\item AT=1,4 cm
\item TF=8,4 cm
\end{itemize}

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

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

 Dans le triangle AZW rectangle en A, on sait que :
\begin{itemize}
\item AZ=2,1 cm
\item \(\widehat{ZWA}\) =  33\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [WZ]. (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 (1.931,2.737) {?};
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 Dans le triangle ABJ rectangle en A, on cherche une relation entre l'angle aigu   \(\widehat{ABJ}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{ABJ}\)) = \(\dfrac{AJ}{AB}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{ABJ}\)) = \(\dfrac{4{,}4}{3{,}3}\)
\par\addvspace{.5em}
 On a donc ABJ = ArcTan(\(\dfrac{4{,}4}{3{,}3}\)) \(\approx\) 53\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{ZTR}\)) = \(\dfrac{ZT}{TR}\)
\par\addvspace{.5em}
d'où cos(64\textdegree{}) = \(\dfrac{ZT}{8{,}7}\)

On a donc ZT = 8,7 \(\times\) cos(64\textdegree{}) \(\approx\) 3,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.290,1.324) {16°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{MSD}\)) = \(\dfrac{MS}{DS}\)
\par\addvspace{.5em}
  d'où    cos(16\textdegree{}) =  \(\dfrac{MS}{1{,}1}\)

On a donc MS = 1,1 \(\times\) cos(16\textdegree{}) \(\approx\) 1{,}1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{AFT}\)) = \(\dfrac{AT}{TF}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{AFT}\)) =\(\dfrac{1{,}4}{8{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{AFT}\) = ArcSin(\(\dfrac{1{,}4}{8{,}4}\)) \(\approx\) 10\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.756,0.467) {W};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.844,1.933) {2,1 cm};
  \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 (5.017,0.723) {33°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{AWZ}\)) = \(\dfrac{AZ}{ZW}\)
\par\addvspace{.5em}
 D'où sin(33\textdegree{}) = \(\dfrac{2{,}1}{ZW}\)
\par\addvspace{.5em}
On a donc ZW =\(\dfrac{2{,}1}{\sin(33^{\circ})}\) \(\approx\) 3,9 cm\end{document}
