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\begin{document}

\sloppy

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

 Dans le triangle RHJ rectangle en R, on sait que :
\begin{itemize}
\item HJ=6,7 cm
\item \(\widehat{RHJ}\) = 52\textdegree{}
\end{itemize}

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

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

 Dans le triangle KFH rectangle en K, on sait que :
\begin{itemize}
\item FH=9,4 cm
\item \(\widehat{KFH}\) = 48\textdegree{}
\end{itemize}

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

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

 Dans le triangle ALM rectangle en A, on sait que :
\begin{itemize}
\item AL=2,1 cm
\item AM=5,3 cm
\end{itemize}

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

 Dans le triangle PNV rectangle en P, on sait que :
\begin{itemize}
\item PV=1,6 cm
\item \(\widehat{PNV}\) = 78\textdegree{}
\end{itemize}

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

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

 Dans le triangle FSC rectangle en F, on sait que :
\begin{itemize}
\item FS=1,8 cm
\item SC=8,4 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{RHJ}\)) =  \(\dfrac{RJ}{HJ}\)
\par\addvspace{.5em}
 d'où sin(52\textdegree{}) = \(\dfrac{RJ}{6{,}7}\)

On a donc RJ = 6,7 \(\times\) sin(52\textdegree{}) \(\approx\) 5,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{KFH}\)) = \(\dfrac{KF}{FH}\)
\par\addvspace{.5em}
d'où cos(48\textdegree{}) = \(\dfrac{KF}{9{,}4}\)

On a donc KF = 9,4 \(\times\) cos(48\textdegree{}) \(\approx\) 6,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{ALM}\)) = \(\dfrac{AM}{AL}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{ALM}\)) = \(\dfrac{5{,}3}{2{,}1}\)
\par\addvspace{.5em}
 On a donc ALM = ArcTan(\(\dfrac{5{,}3}{2{,}1}\)) \(\approx\) 68\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{PNV}\)) =  \(\dfrac{PV}{NV}\)
\par\addvspace{.5em}
 d'où sin(78\textdegree{}) = \(\dfrac{1{,}6}{NV}\)
\par\addvspace{.5em}
 On a donc NV = \(\dfrac{1{,}6}{\sin(78^{\circ})}\) \(\approx\) 1,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.287,1.468) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{FCS}\)) = \(\dfrac{FS}{SC}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{FCS}\)) =\(\dfrac{1{,}8}{8{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{FCS}\) = ArcSin(\(\dfrac{1{,}8}{8{,}4}\)) \(\approx\) 12\textdegree{}\end{document}
