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\begin{document}

\sloppy

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

 Dans le triangle ANL rectangle en A, on sait que :
\begin{itemize}
\item AL=9,3 cm
\item \(\widehat{ANL}\) = 77\textdegree{}
\end{itemize}

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

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

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

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

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

 Dans le triangle BKW rectangle en B, on sait que :
\begin{itemize}
\item KW=9,1 cm
\item \(\widehat{BKW}\) = 80\textdegree{}
\end{itemize}

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

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

 Dans le triangle CPN rectangle en C, on sait que :
\begin{itemize}
\item CP=1,6 cm
\item PN=8,8 cm
\end{itemize}

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

 Dans le triangle RDT rectangle en R, on sait que :
\begin{itemize}
\item RD=1,8 cm
\item DT=7,8 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{ANL}\)) =  \(\dfrac{AL}{NL}\)
\par\addvspace{.5em}
 d'où sin(77\textdegree{}) = \(\dfrac{9{,}3}{NL}\)
\par\addvspace{.5em}
 On a donc NL = \(\dfrac{9{,}3}{\sin(77^{\circ})}\) \(\approx\) 9,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{KRT}\)) = \(\dfrac{KT}{KR}\)
\par\addvspace{.5em}
d'où tan(77\textdegree{}) =\(\dfrac{KT}{6{,}6}\)
\par\addvspace{.5em}
On a donc KT = 6,6 \(\times\) tan(77\textdegree{}) \(\approx\) 28,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{BKW}\)) = \(\dfrac{BK}{KW}\)
\par\addvspace{.5em}
d'où cos(80\textdegree{}) = \(\dfrac{BK}{9{,}1}\)

On a donc BK = 9,1 \(\times\) cos(80\textdegree{}) \(\approx\) 1,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{CPN}\)) = \(\dfrac{CP}{PN}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{CPN}\)) = \(\dfrac{1{,}6}{8{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{CPN}\) =ArcCos(\(\dfrac{1{,}6}{8{,}8}\)) \(\approx\)80\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.288,1.430) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{RTD}\)) = \(\dfrac{RD}{DT}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{RTD}\)) =\(\dfrac{1{,}8}{7{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{RTD}\) = ArcSin(\(\dfrac{1{,}8}{7{,}8}\)) \(\approx\) 13\textdegree{}\end{document}
