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\begin{document}

\sloppy

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

 Dans le triangle JBL rectangle en J, on sait que :
\begin{itemize}
\item BL=2,5 cm
\item \(\widehat{BLJ}\) =  19\textdegree{}
\end{itemize}

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

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

 Dans le triangle TGM rectangle en T, on sait que :
\begin{itemize}
\item TM=3,6 cm
\item \(\widehat{TGM}\) = 69\textdegree{}
\end{itemize}

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

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

 Dans le triangle WAJ rectangle en W, on sait que :
\begin{itemize}
\item WA=2 cm
\item WJ=5,1 cm
\end{itemize}

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

 Dans le triangle TSR rectangle en T, on sait que :
\begin{itemize}
\item SR=5,9 cm
\item \(\widehat{SRT}\) =  30\textdegree{}
\end{itemize}

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

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

 Dans le triangle TRK rectangle en T, on sait que :
\begin{itemize}
\item TR=1,9 cm
\item TK=5,8 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{JLB}\)) = \(\dfrac{JB}{BL}\)
\par\addvspace{.5em}
 D'où sin(19\textdegree{}) = \(\dfrac{JB}{2{,}5}\)

On a donc JB = 2,5 \(\times\) sin(19\textdegree{}) \(\approx\) 0,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{TGM}\)) =  \(\dfrac{TM}{GM}\)
\par\addvspace{.5em}
 d'où sin(69\textdegree{}) = \(\dfrac{3{,}6}{GM}\)
\par\addvspace{.5em}
 On a donc GM = \(\dfrac{3{,}6}{\sin(69^{\circ})}\) \(\approx\) 3,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{WAJ}\)) = \(\dfrac{WJ}{WA}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{WAJ}\)) = \(\dfrac{5{,}1}{2}\)
\par\addvspace{.5em}
 On a donc WAJ = ArcTan(\(\dfrac{5{,}1}{2}\)) \(\approx\) 69\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{TRS}\)) = \(\dfrac{TR}{SR}\)
\par\addvspace{.5em}
  d'où    cos(30\textdegree{}) =  \(\dfrac{TR}{5{,}9}\)

On a donc TR = 5,9 \(\times\) cos(30\textdegree{}) \(\approx\) 5{,}1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.864) {5,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.292,1.236) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{TKR}\)) = \(\dfrac{TR}{TK}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{TKR}\)) =  \(\dfrac{1{,}9}{5{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{TKR}\) =  ArcTan(\(\dfrac{1{,}9}{5{,}8}\)) \(\approx\) 18\textdegree{}\end{document}
