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\begin{document}

\sloppy

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

 Dans le triangle DWR rectangle en D, on sait que :
\begin{itemize}
\item DR=10 cm
\item \(\widehat{WRD}\) =  36\textdegree{}
\end{itemize}

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

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

 Dans le triangle HMW rectangle en H, on sait que :
\begin{itemize}
\item HM=6,6 cm
\item \(\widehat{MWH}\) =  44\textdegree{}
\end{itemize}

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

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

 Dans le triangle FST rectangle en F, on sait que :
\begin{itemize}
\item FS=2,9 cm
\item FT=6 cm
\end{itemize}

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

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

 Dans le triangle WCF rectangle en W, on sait que :
\begin{itemize}
\item WF=4,3 cm
\item CF=8,4 cm
\end{itemize}

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

 Dans le triangle NML rectangle en N, on sait que :
\begin{itemize}
\item NM=5,1 cm
\item \(\widehat{MLN}\) =  10\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{DRW}\)) = \(\dfrac{DW}{DR}\)
\par\addvspace{.5em}
d'où  tan(36\textdegree{}) =  \(\dfrac{DW}{10}\)
\par\addvspace{.5em}
On a donc DW =  10 \(\times\) tan(36\textdegree{})  \(\approx\) 7,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{HWM}\)) = \(\dfrac{HM}{MW}\)
\par\addvspace{.5em}
 D'où sin(44\textdegree{}) = \(\dfrac{6{,}6}{MW}\)
\par\addvspace{.5em}
On a donc MW =\(\dfrac{6{,}6}{\sin(44^{\circ})}\) \(\approx\) 9,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{FTS}\)) = \(\dfrac{FS}{FT}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{FTS}\)) =  \(\dfrac{2{,}9}{6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{FTS}\) =  ArcTan(\(\dfrac{2{,}9}{6}\)) \(\approx\) 26\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{WCF}\)) =  \(\dfrac{WF}{CF}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{WCF}\)) = \(\dfrac{4{,}3}{8{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{WCF}\) =  ArcSin(\(\dfrac{4{,}3}{8{,}4}\)) \(\approx\) 31\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.286,1.559) {10°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{NLM}\)) = \(\dfrac{NM}{NL}\)
\par\addvspace{.5em}
d'où  tan(10\textdegree{}) =  \(\dfrac{5{,}1}{NL}\)
\par\addvspace{.5em}
On a donc NL = \(\dfrac{5{,}1}{\tan(10^{\circ})}\) \(\approx\) 28,9 cm\end{document}
