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\begin{document}

\sloppy

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

 Dans le triangle BRJ rectangle en B, on sait que :
\begin{itemize}
\item BR=2,8 cm
\item RJ=7,6 cm
\end{itemize}

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

 Dans le triangle ADK rectangle en A, on sait que :
\begin{itemize}
\item AK=5,8 cm
\item DK=10 cm
\end{itemize}

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

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

 Dans le triangle RDZ rectangle en R, on sait que :
\begin{itemize}
\item RD=8,8 cm
\item \(\widehat{DZR}\) =  20\textdegree{}
\end{itemize}

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

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

 Dans le triangle WGH rectangle en W, on sait que :
\begin{itemize}
\item WG=4,9 cm
\item \(\widehat{WGH}\) = 62\textdegree{}
\end{itemize}

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

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

 Dans le triangle LJA rectangle en L, on sait que :
\begin{itemize}
\item JA=3,4 cm
\item \(\widehat{JAL}\) =  39\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{BRJ}\)) = \(\dfrac{BR}{RJ}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{BRJ}\)) = \(\dfrac{2{,}8}{7{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{BRJ}\) =ArcCos(\(\dfrac{2{,}8}{7{,}6}\)) \(\approx\)68\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{AKD}\)) = \(\dfrac{AK}{DK}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{AKD}\)) =  \(\dfrac{5{,}8}{10}\)
\par\addvspace{.5em}
On a donc \(\widehat{AKD}\) = Arccos (\(\dfrac{5{,}8}{10}\)) \(\approx\) 55\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{RZD}\)) = \(\dfrac{RD}{DZ}\)
\par\addvspace{.5em}
 D'où sin(20\textdegree{}) = \(\dfrac{8{,}8}{DZ}\)
\par\addvspace{.5em}
On a donc DZ =\(\dfrac{8{,}8}{\sin(20^{\circ})}\) \(\approx\) 25,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{WGH}\)) = \(\dfrac{WH}{WG}\)
\par\addvspace{.5em}
d'où tan(62\textdegree{}) =\(\dfrac{WH}{4{,}9}\)
\par\addvspace{.5em}
On a donc WH = 4,9 \(\times\) tan(62\textdegree{}) \(\approx\) 9,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.600,0.756) {39°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{LAJ}\)) = \(\dfrac{LJ}{JA}\)
\par\addvspace{.5em}
 D'où sin(39\textdegree{}) = \(\dfrac{LJ}{3{,}4}\)

On a donc LJ = 3,4 \(\times\) sin(39\textdegree{}) \(\approx\) 2,1 cm\end{document}
