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\begin{document}

\sloppy

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

 Dans le triangle KMN rectangle en K, on sait que :
\begin{itemize}
\item KM=1 cm
\item MN=7 cm
\end{itemize}

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

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

 Dans le triangle VML rectangle en V, on sait que :
\begin{itemize}
\item VL=3,1 cm
\item \(\widehat{VML}\) = 47\textdegree{}
\end{itemize}

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

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

 Dans le triangle RHW rectangle en R, on sait que :
\begin{itemize}
\item RW=4,1 cm
\item \(\widehat{RHW}\) = 63\textdegree{}
\end{itemize}

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

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

 Dans le triangle BSK rectangle en B, on sait que :
\begin{itemize}
\item SK=2,8 cm
\item \(\widehat{BSK}\) = 61\textdegree{}
\end{itemize}

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

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

 Dans le triangle MBN rectangle en M, on sait que :
\begin{itemize}
\item MB=1,4 cm
\item MN=5,6 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{MBN}\). \newpage
\section*{\textcolor{red}{Correction}}

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

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

\par\addvspace{.5em}
sin(\(\widehat{KNM}\)) = \(\dfrac{KM}{MN}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{KNM}\)) =\(\dfrac{1}{7}\)
\par\addvspace{.5em}
 On a donc \(\widehat{KNM}\) = ArcSin(\(\dfrac{1}{7}\)) \(\approx\) 8\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{VML}\)) =  \(\dfrac{VL}{ML}\)
\par\addvspace{.5em}
 d'où sin(47\textdegree{}) = \(\dfrac{3{,}1}{ML}\)
\par\addvspace{.5em}
 On a donc ML = \(\dfrac{3{,}1}{\sin(47^{\circ})}\) \(\approx\) 4,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{RHW}\)) = \(\dfrac{RW}{RH}\)
\par\addvspace{.5em}
tan(63\textdegree{}) = \(\dfrac{4{,}1}{RH}\)
\par\addvspace{.5em}
 On a donc RH = \(\dfrac{4{,}1}{\tan(63^{\circ})}\) \(\approx\) 2,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{BSK}\)) =  \(\dfrac{BK}{SK}\)
\par\addvspace{.5em}
 d'où sin(61\textdegree{}) = \(\dfrac{BK}{2{,}8}\)

On a donc BK = 2,8 \(\times\) sin(61\textdegree{}) \(\approx\) 2,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.460,2.020) {?};
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 Dans le triangle MBN rectangle en M, on cherche une relation entre l'angle aigu   \(\widehat{MBN}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{MBN}\)) = \(\dfrac{MN}{MB}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{MBN}\)) = \(\dfrac{5{,}6}{1{,}4}\)
\par\addvspace{.5em}
 On a donc MBN = ArcTan(\(\dfrac{5{,}6}{1{,}4}\)) \(\approx\) 76\textdegree{}\end{document}
