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\begin{document}

\sloppy

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

 Dans le triangle LJG rectangle en L, on sait que :
\begin{itemize}
\item JG=7,9 cm
\item \(\widehat{JGL}\) =  38\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZWL rectangle en Z, on sait que :
\begin{itemize}
\item WL=7,1 cm
\item \(\widehat{WLZ}\) =  27\textdegree{}
\end{itemize}

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

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

 Dans le triangle FKL rectangle en F, on sait que :
\begin{itemize}
\item FK=2,2 cm
\item FL=4,8 cm
\end{itemize}

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

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

 Dans le triangle BVS rectangle en B, on sait que :
\begin{itemize}
\item BV=1,2 cm
\item VS=9,5 cm
\end{itemize}

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

 Dans le triangle RBJ rectangle en R, on sait que :
\begin{itemize}
\item RJ=8,2 cm
\item \(\widehat{RBJ}\) = 75\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{LGJ}\)) = \(\dfrac{LJ}{JG}\)
\par\addvspace{.5em}
 D'où sin(38\textdegree{}) = \(\dfrac{LJ}{7{,}9}\)

On a donc LJ = 7,9 \(\times\) sin(38\textdegree{}) \(\approx\) 4,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{ZLW}\)) = \(\dfrac{ZL}{WL}\)
\par\addvspace{.5em}
  d'où    cos(27\textdegree{}) =  \(\dfrac{ZL}{7{,}1}\)

On a donc ZL = 7,1 \(\times\) cos(27\textdegree{}) \(\approx\) 6{,}3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{FLK}\)) = \(\dfrac{FK}{FL}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{FLK}\)) =  \(\dfrac{2{,}2}{4{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{FLK}\) =  ArcTan(\(\dfrac{2{,}2}{4{,}8}\)) \(\approx\) 25\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{BVS}\)) = \(\dfrac{BV}{VS}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{BVS}\)) = \(\dfrac{1{,}2}{9{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{BVS}\) =ArcCos(\(\dfrac{1{,}2}{9{,}5}\)) \(\approx\)83\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.010) {8,2 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 (1.456,2.061) {75°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{RBJ}\)) =  \(\dfrac{RJ}{BJ}\)
\par\addvspace{.5em}
 d'où sin(75\textdegree{}) = \(\dfrac{8{,}2}{BJ}\)
\par\addvspace{.5em}
 On a donc BJ = \(\dfrac{8{,}2}{\sin(75^{\circ})}\) \(\approx\) 8,5 cm\end{document}
