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\begin{document}

\sloppy

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

 Dans le triangle BRL rectangle en B, on sait que :
\begin{itemize}
\item RL=6,4 cm
\item \(\widehat{RLB}\) =  37\textdegree{}
\end{itemize}

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

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

 Dans le triangle KSL rectangle en K, on sait que :
\begin{itemize}
\item SL=9,9 cm
\item \(\widehat{SLK}\) =  29\textdegree{}
\end{itemize}

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

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

 Dans le triangle CAV rectangle en C, on sait que :
\begin{itemize}
\item CV=4,1 cm
\item \(\widehat{AVC}\) =  34\textdegree{}
\end{itemize}

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

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

 Dans le triangle GHJ rectangle en G, on sait que :
\begin{itemize}
\item GH=2,9 cm
\item GJ=5 cm
\end{itemize}

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

 Dans le triangle VNK rectangle en V, on sait que :
\begin{itemize}
\item VK=4,1 cm
\item NK=8,4 cm
\end{itemize}

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

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

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

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.726,0.745) {37°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{BLR}\)) = \(\dfrac{BL}{RL}\)
\par\addvspace{.5em}
  d'où    cos(37\textdegree{}) =  \(\dfrac{BL}{6{,}4}\)

On a donc BL = 6,4 \(\times\) cos(37\textdegree{}) \(\approx\) 5{,}1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{KLS}\)) = \(\dfrac{KS}{SL}\)
\par\addvspace{.5em}
 D'où sin(29\textdegree{}) = \(\dfrac{KS}{9{,}9}\)

On a donc KS = 9,9 \(\times\) sin(29\textdegree{}) \(\approx\) 4,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{CVA}\)) = \(\dfrac{CV}{AV}\)
\par\addvspace{.5em}
  d'où    cos(34\textdegree{}) =  \(\dfrac{4{,}1}{AV}\)
\par\addvspace{.5em}
 On a donc AV =  \(\dfrac{4{,}1}{\cos(34^{\circ})}\) \(\approx\) 4,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{GHJ}\)) = \(\dfrac{GJ}{GH}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{GHJ}\)) = \(\dfrac{5}{2{,}9}\)
\par\addvspace{.5em}
 On a donc GHJ = ArcTan(\(\dfrac{5}{2{,}9}\)) \(\approx\) 60\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {4,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.708,0.871) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{VKN}\)) = \(\dfrac{VK}{NK}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{VKN}\)) =  \(\dfrac{4{,}1}{8{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{VKN}\) = Arccos (\(\dfrac{4{,}1}{8{,}4}\)) \(\approx\) 61\textdegree{} \end{document}
