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\begin{document}

\sloppy

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

 Dans le triangle DRA rectangle en D, on sait que :
\begin{itemize}
\item RA=5,1 cm
\item \(\widehat{RAD}\) =  22\textdegree{}
\end{itemize}

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

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

 Dans le triangle BJR rectangle en B, on sait que :
\begin{itemize}
\item BR=3,5 cm
\item JR=8,4 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{BJR}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle WDC rectangle en W, on sait que :
\begin{itemize}
\item WD=1 cm
\item WC=5,8 cm
\end{itemize}

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

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

 Dans le triangle DCA rectangle en D, on sait que :
\begin{itemize}
\item DA=8,4 cm
\item \(\widehat{DCA}\) = 66\textdegree{}
\end{itemize}

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

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

 Dans le triangle GTW rectangle en G, on sait que :
\begin{itemize}
\item GT=1,1 cm
\item \(\widehat{GTW}\) = 75\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{DAR}\)) = \(\dfrac{DA}{RA}\)
\par\addvspace{.5em}
  d'où    cos(22\textdegree{}) =  \(\dfrac{DA}{5{,}1}\)

On a donc DA = 5,1 \(\times\) cos(22\textdegree{}) \(\approx\) 4{,}7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{BJR}\)) =  \(\dfrac{BR}{JR}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{BJR}\)) = \(\dfrac{3{,}5}{8{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{BJR}\) =  ArcSin(\(\dfrac{3{,}5}{8{,}4}\)) \(\approx\) 25\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{WCD}\)) = \(\dfrac{WD}{WC}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{WCD}\)) =  \(\dfrac{1}{5{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{WCD}\) =  ArcTan(\(\dfrac{1}{5{,}8}\)) \(\approx\) 10\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{DCA}\)) = \(\dfrac{DA}{DC}\)
\par\addvspace{.5em}
tan(66\textdegree{}) = \(\dfrac{8{,}4}{DC}\)
\par\addvspace{.5em}
 On a donc DC = \(\dfrac{8{,}4}{\tan(66^{\circ})}\) \(\approx\) 3,7 cm
\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 (1.456,2.061) {75°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{GTW}\)) = \(\dfrac{GT}{TW}\)
\par\addvspace{.5em}
 d'où cos(75\textdegree{}) = \(\dfrac{1{,}1}{TW}\)
\par\addvspace{.5em}
On a donc TW = \(\dfrac{1{,}1}{\cos(75^{\circ})}\)\(\approx\) 4,3 cm\end{document}
