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\begin{document}

\sloppy

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

 Dans le triangle SJP rectangle en S, on sait que :
\begin{itemize}
\item SJ=6,8 cm
\item \(\widehat{JPS}\) =  20\textdegree{}
\end{itemize}

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

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

 Dans le triangle GCJ rectangle en G, on sait que :
\begin{itemize}
\item GJ=4,3 cm
\item CJ=8,7 cm
\end{itemize}

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

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

 Dans le triangle LTM rectangle en L, on sait que :
\begin{itemize}
\item TM=7,7 cm
\item \(\widehat{TML}\) =  40\textdegree{}
\end{itemize}

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

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

 Dans le triangle SBK rectangle en S, on sait que :
\begin{itemize}
\item SB=2,5 cm
\item BK=7 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{SBK}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle CTV rectangle en C, on sait que :
\begin{itemize}
\item CT=6,4 cm
\item \(\widehat{TVC}\) =  36\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{SPJ}\)) = \(\dfrac{SJ}{JP}\)
\par\addvspace{.5em}
 D'où sin(20\textdegree{}) = \(\dfrac{6{,}8}{JP}\)
\par\addvspace{.5em}
On a donc JP =\(\dfrac{6{,}8}{\sin(20^{\circ})}\) \(\approx\) 19,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{GJC}\)) = \(\dfrac{GJ}{CJ}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{GJC}\)) =  \(\dfrac{4{,}3}{8{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{GJC}\) = Arccos (\(\dfrac{4{,}3}{8{,}7}\)) \(\approx\) 60\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{LMT}\)) = \(\dfrac{LT}{TM}\)
\par\addvspace{.5em}
 D'où sin(40\textdegree{}) = \(\dfrac{LT}{7{,}7}\)

On a donc LT = 7,7 \(\times\) sin(40\textdegree{}) \(\approx\) 4,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{SBK}\)) = \(\dfrac{SB}{BK}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{SBK}\)) = \(\dfrac{2{,}5}{7}\)
\par\addvspace{.5em}
On a donc \(\widehat{SBK}\) =ArcCos(\(\dfrac{2{,}5}{7}\)) \(\approx\)69\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.793,0.739) {36°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{CVT}\)) = \(\dfrac{CT}{CV}\)
\par\addvspace{.5em}
d'où  tan(36\textdegree{}) =  \(\dfrac{6{,}4}{CV}\)
\par\addvspace{.5em}
On a donc CV = \(\dfrac{6{,}4}{\tan(36^{\circ})}\) \(\approx\) 8,8 cm\end{document}
