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\begin{document}

\sloppy

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

 Dans le triangle KNB rectangle en K, on sait que :
\begin{itemize}
\item NB=7,6 cm
\item \(\widehat{NBK}\) =  36\textdegree{}
\end{itemize}

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

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

 Dans le triangle BCL rectangle en B, on sait que :
\begin{itemize}
\item BL=6,5 cm
\item CL=7,6 cm
\end{itemize}

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

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

 Dans le triangle JLG rectangle en J, on sait que :
\begin{itemize}
\item JL=3 cm
\item JG=3,7 cm
\end{itemize}

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

 Dans le triangle NRF rectangle en N, on sait que :
\begin{itemize}
\item NR=5 cm
\item \(\widehat{RFN}\) =  41\textdegree{}
\end{itemize}

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

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

 Dans le triangle KML rectangle en K, on sait que :
\begin{itemize}
\item KL=2,6 cm
\item \(\widehat{KML}\) = 77\textdegree{}
\end{itemize}

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

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

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

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{KBN}\)) = \(\dfrac{KB}{NB}\)
\par\addvspace{.5em}
  d'où    cos(36\textdegree{}) =  \(\dfrac{KB}{7{,}6}\)

On a donc KB = 7,6 \(\times\) cos(36\textdegree{}) \(\approx\) 6{,}1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{BLC}\)) = \(\dfrac{BL}{CL}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{BLC}\)) =  \(\dfrac{6{,}5}{7{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{BLC}\) = Arccos (\(\dfrac{6{,}5}{7{,}6}\)) \(\approx\) 31\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{JLG}\)) = \(\dfrac{JG}{JL}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{JLG}\)) = \(\dfrac{3{,}7}{3}\)
\par\addvspace{.5em}
 On a donc JLG = ArcTan(\(\dfrac{3{,}7}{3}\)) \(\approx\) 51\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.486,0.767) {41°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{NFR}\)) = \(\dfrac{NR}{RF}\)
\par\addvspace{.5em}
 D'où sin(41\textdegree{}) = \(\dfrac{5}{RF}\)
\par\addvspace{.5em}
On a donc RF =\(\dfrac{5}{\sin(41^{\circ})}\) \(\approx\) 7,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.465,1.977) {77°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{KML}\)) = \(\dfrac{KL}{KM}\)
\par\addvspace{.5em}
tan(77\textdegree{}) = \(\dfrac{2{,}6}{KM}\)
\par\addvspace{.5em}
 On a donc KM = \(\dfrac{2{,}6}{\tan(77^{\circ})}\) \(\approx\) 0,6 cm\end{document}
