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\begin{document}

\sloppy

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

 Dans le triangle JHS rectangle en J, on sait que :
\begin{itemize}
\item JS=5,2 cm
\item HS=6,9 cm
\end{itemize}

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

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

 Dans le triangle SNB rectangle en S, on sait que :
\begin{itemize}
\item SB=7,4 cm
\item \(\widehat{NBS}\) =  43\textdegree{}
\end{itemize}

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

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

 Dans le triangle MFK rectangle en M, on sait que :
\begin{itemize}
\item MK=8,2 cm
\item \(\widehat{MFK}\) = 60\textdegree{}
\end{itemize}

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

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

 Dans le triangle NLF rectangle en N, on sait que :
\begin{itemize}
\item NL=1,2 cm
\item LF=7,8 cm
\end{itemize}

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

 Dans le triangle TRA rectangle en T, on sait que :
\begin{itemize}
\item RA=5 cm
\item \(\widehat{RAT}\) =  21\textdegree{}
\end{itemize}

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

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

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

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.481,0.767) {?};
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 Dans le triangle JHS rectangle en J, on cherche une relation entre l'angle aigu   \(\widehat{JSH}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{JSH}\)) = \(\dfrac{JS}{HS}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{JSH}\)) =  \(\dfrac{5{,}2}{6{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{JSH}\) = Arccos (\(\dfrac{5{,}2}{6{,}9}\)) \(\approx\) 41\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{SBN}\)) = \(\dfrac{SN}{SB}\)
\par\addvspace{.5em}
d'où  tan(43\textdegree{}) =  \(\dfrac{SN}{7{,}4}\)
\par\addvspace{.5em}
On a donc SN =  7,4 \(\times\) tan(43\textdegree{})  \(\approx\) 6,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{MFK}\)) =  \(\dfrac{MK}{FK}\)
\par\addvspace{.5em}
 d'où sin(60\textdegree{}) = \(\dfrac{8{,}2}{FK}\)
\par\addvspace{.5em}
 On a donc FK = \(\dfrac{8{,}2}{\sin(60^{\circ})}\) \(\approx\) 9,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{NLF}\)) = \(\dfrac{NL}{LF}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{NLF}\)) = \(\dfrac{1{,}2}{7{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{NLF}\) =ArcCos(\(\dfrac{1{,}2}{7{,}8}\)) \(\approx\)81\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

cos(\(\widehat{TAR}\)) = \(\dfrac{TA}{RA}\)
\par\addvspace{.5em}
  d'où    cos(21\textdegree{}) =  \(\dfrac{TA}{5}\)

On a donc TA = 5 \(\times\) cos(21\textdegree{}) \(\approx\) 4{,}7 cm\end{document}
