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\begin{document}

\sloppy

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

 Dans le triangle MKB rectangle en M, on sait que :
\begin{itemize}
\item MK=2,3 cm
\item KB=8,8 cm
\end{itemize}

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

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

 Dans le triangle GJA rectangle en G, on sait que :
\begin{itemize}
\item JA=4,8 cm
\item \(\widehat{JAG}\) =  27\textdegree{}
\end{itemize}

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

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

 Dans le triangle VMB rectangle en V, on sait que :
\begin{itemize}
\item VM=1,6 cm
\item MB=7,2 cm
\end{itemize}

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

 Dans le triangle NGR rectangle en N, on sait que :
\begin{itemize}
\item NR=3,4 cm
\item \(\widehat{NGR}\) = 46\textdegree{}
\end{itemize}

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

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

 Dans le triangle CBL rectangle en C, on sait que :
\begin{itemize}
\item CL=1,6 cm
\item \(\widehat{CBL}\) = 67\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{MBK}\)) = \(\dfrac{MK}{KB}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{MBK}\)) =\(\dfrac{2{,}3}{8{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{MBK}\) = ArcSin(\(\dfrac{2{,}3}{8{,}8}\)) \(\approx\) 15\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{GAJ}\)) = \(\dfrac{GA}{JA}\)
\par\addvspace{.5em}
  d'où    cos(27\textdegree{}) =  \(\dfrac{GA}{4{,}8}\)

On a donc GA = 4,8 \(\times\) cos(27\textdegree{}) \(\approx\) 4{,}3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{VMB}\)) = \(\dfrac{VM}{MB}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{VMB}\)) = \(\dfrac{1{,}6}{7{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{VMB}\) =ArcCos(\(\dfrac{1{,}6}{7{,}2}\)) \(\approx\)77\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{NGR}\)) = \(\dfrac{NR}{NG}\)
\par\addvspace{.5em}
tan(46\textdegree{}) = \(\dfrac{3{,}4}{NG}\)
\par\addvspace{.5em}
 On a donc NG = \(\dfrac{3{,}4}{\tan(46^{\circ})}\) \(\approx\) 3,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \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 (1.418,2.417) {67°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{CBL}\)) =  \(\dfrac{CL}{BL}\)
\par\addvspace{.5em}
 d'où sin(67\textdegree{}) = \(\dfrac{1{,}6}{BL}\)
\par\addvspace{.5em}
 On a donc BL = \(\dfrac{1{,}6}{\sin(67^{\circ})}\) \(\approx\) 1,7 cm\end{document}
