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\begin{document}

\sloppy

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

 Dans le triangle TNL rectangle en T, on sait que :
\begin{itemize}
\item TL=3,1 cm
\item \(\widehat{TNL}\) = 67\textdegree{}
\end{itemize}

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

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

 Dans le triangle DFJ rectangle en D, on sait que :
\begin{itemize}
\item FJ=7,8 cm
\item \(\widehat{DFJ}\) = 68\textdegree{}
\end{itemize}

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

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

 Dans le triangle HAS rectangle en H, on sait que :
\begin{itemize}
\item HA=1,5 cm
\item AS=9,6 cm
\end{itemize}

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

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

 Dans le triangle NKT rectangle en N, on sait que :
\begin{itemize}
\item NT=4,2 cm
\item KT=9,1 cm
\end{itemize}

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

 Dans le triangle SHM rectangle en S, on sait que :
\begin{itemize}
\item SM=7,4 cm
\item \(\widehat{SHM}\) = 52\textdegree{}
\end{itemize}

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

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

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

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 Dans le triangle TNL rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TNL}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{TNL}\)) = \(\dfrac{TL}{TN}\)
\par\addvspace{.5em}
tan(67\textdegree{}) = \(\dfrac{3{,}1}{TN}\)
\par\addvspace{.5em}
 On a donc TN = \(\dfrac{3{,}1}{\tan(67^{\circ})}\) \(\approx\) 1,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{DFJ}\)) =  \(\dfrac{DJ}{FJ}\)
\par\addvspace{.5em}
 d'où sin(68\textdegree{}) = \(\dfrac{DJ}{7{,}8}\)

On a donc DJ = 7,8 \(\times\) sin(68\textdegree{}) \(\approx\) 7,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{HSA}\)) = \(\dfrac{HA}{AS}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{HSA}\)) =\(\dfrac{1{,}5}{9{,}6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{HSA}\) = ArcSin(\(\dfrac{1{,}5}{9{,}6}\)) \(\approx\) 9\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{NKT}\)) =  \(\dfrac{NT}{KT}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{NKT}\)) = \(\dfrac{4{,}2}{9{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{NKT}\) =  ArcSin(\(\dfrac{4{,}2}{9{,}1}\)) \(\approx\) 27\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{SHM}\)) =  \(\dfrac{SM}{HM}\)
\par\addvspace{.5em}
 d'où sin(52\textdegree{}) = \(\dfrac{7{,}4}{HM}\)
\par\addvspace{.5em}
 On a donc HM = \(\dfrac{7{,}4}{\sin(52^{\circ})}\) \(\approx\) 9,4 cm\end{document}
