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\begin{document}

\sloppy

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

 Dans le triangle JHR rectangle en J, on sait que :
\begin{itemize}
\item JH=5,1 cm
\item \(\widehat{HRJ}\) =  32\textdegree{}
\end{itemize}

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

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

 Dans le triangle PHC rectangle en P, on sait que :
\begin{itemize}
\item PH=1 cm
\item PC=3,7 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{PHC}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle MCJ rectangle en M, on sait que :
\begin{itemize}
\item CJ=8 cm
\item \(\widehat{CJM}\) =  12\textdegree{}
\end{itemize}

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

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

 Dans le triangle VHS rectangle en V, on sait que :
\begin{itemize}
\item HS=1 cm
\item \(\widehat{HSV}\) =  11\textdegree{}
\end{itemize}

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

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

 Dans le triangle VCD rectangle en V, on sait que :
\begin{itemize}
\item VC=2,5 cm
\item VD=3,9 cm
\end{itemize}

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

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

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

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{JRH}\)) = \(\dfrac{JH}{HR}\)
\par\addvspace{.5em}
 D'où sin(32\textdegree{}) = \(\dfrac{5{,}1}{HR}\)
\par\addvspace{.5em}
On a donc HR =\(\dfrac{5{,}1}{\sin(32^{\circ})}\) \(\approx\) 9,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{PHC}\)) = \(\dfrac{PC}{PH}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{PHC}\)) = \(\dfrac{3{,}7}{1}\)
\par\addvspace{.5em}
 On a donc PHC = ArcTan(3,7) \(\approx\) 75\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{MJC}\)) = \(\dfrac{MC}{CJ}\)
\par\addvspace{.5em}
 D'où sin(12\textdegree{}) = \(\dfrac{MC}{8}\)

On a donc MC = 8 \(\times\) sin(12\textdegree{}) \(\approx\) 1,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{VSH}\)) = \(\dfrac{VS}{HS}\)
\par\addvspace{.5em}
  d'où    cos(11\textdegree{}) =  \(\dfrac{VS}{1}\)

On a donc VS = 1 \(\times\) cos(11\textdegree{}) \(\approx\) 1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.044,0.721) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{VDC}\)) = \(\dfrac{VC}{VD}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{VDC}\)) =  \(\dfrac{2{,}5}{3{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{VDC}\) =  ArcTan(\(\dfrac{2{,}5}{3{,}9}\)) \(\approx\) 33\textdegree{}\end{document}
