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\begin{document}

\sloppy

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

 Dans le triangle DHF rectangle en D, on sait que :
\begin{itemize}
\item DH=7,3 cm
\item \(\widehat{HFD}\) =  23\textdegree{}
\end{itemize}

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

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

 Dans le triangle MLB rectangle en M, on sait que :
\begin{itemize}
\item ML=0,7 cm
\item \(\widehat{MLB}\) = 47\textdegree{}
\end{itemize}

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

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

 Dans le triangle ANF rectangle en A, on sait que :
\begin{itemize}
\item AF=6,5 cm
\item NF=8,7 cm
\end{itemize}

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

 Dans le triangle NMS rectangle en N, on sait que :
\begin{itemize}
\item NS=5 cm
\item MS=7,8 cm
\end{itemize}

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

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

 Dans le triangle JMR rectangle en J, on sait que :
\begin{itemize}
\item MR=6,6 cm
\item \(\widehat{MRJ}\) =  42\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{DFH}\)) = \(\dfrac{DH}{DF}\)
\par\addvspace{.5em}
d'où  tan(23\textdegree{}) =  \(\dfrac{7{,}3}{DF}\)
\par\addvspace{.5em}
On a donc DF = \(\dfrac{7{,}3}{\tan(23^{\circ})}\) \(\approx\) 17,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{MLB}\)) = \(\dfrac{ML}{LB}\)
\par\addvspace{.5em}
 d'où cos(47\textdegree{}) = \(\dfrac{0{,}7}{LB}\)
\par\addvspace{.5em}
On a donc LB = \(\dfrac{0{,}7}{\cos(47^{\circ})}\)\(\approx\) 1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{ANF}\)) =  \(\dfrac{AF}{NF}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{ANF}\)) = \(\dfrac{6{,}5}{8{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{ANF}\) =  ArcSin(\(\dfrac{6{,}5}{8{,}7}\)) \(\approx\) 48\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{NSM}\)) = \(\dfrac{NS}{MS}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{NSM}\)) =  \(\dfrac{5}{7{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{NSM}\) = Arccos (\(\dfrac{5}{7{,}8}\)) \(\approx\) 50\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{JRM}\)) = \(\dfrac{JM}{MR}\)
\par\addvspace{.5em}
 D'où sin(42\textdegree{}) = \(\dfrac{JM}{6{,}6}\)

On a donc JM = 6,6 \(\times\) sin(42\textdegree{}) \(\approx\) 4,4 cm\end{document}
