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\begin{document}

\sloppy

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

 Dans le triangle DPN rectangle en D, on sait que :
\begin{itemize}
\item DP=2 cm
\item DN=4,5 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{DPN}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle SHT rectangle en S, on sait que :
\begin{itemize}
\item HT=0,5 cm
\item \(\widehat{SHT}\) = 50\textdegree{}
\end{itemize}

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

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

 Dans le triangle CGM rectangle en C, on sait que :
\begin{itemize}
\item GM=4,8 cm
\item \(\widehat{CGM}\) = 75\textdegree{}
\end{itemize}

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

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

 Dans le triangle DVT rectangle en D, on sait que :
\begin{itemize}
\item DT=0,9 cm
\item \(\widehat{DVT}\) = 60\textdegree{}
\end{itemize}

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

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

 Dans le triangle HPK rectangle en H, on sait que :
\begin{itemize}
\item HP=1,7 cm
\item HK=4,7 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{DPN}\)) = \(\dfrac{DN}{DP}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{DPN}\)) = \(\dfrac{4{,}5}{2}\)
\par\addvspace{.5em}
 On a donc DPN = ArcTan(\(\dfrac{4{,}5}{2}\)) \(\approx\) 66\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{SHT}\)) = \(\dfrac{SH}{HT}\)
\par\addvspace{.5em}
d'où cos(50\textdegree{}) = \(\dfrac{SH}{0{,}5}\)

On a donc SH = 0,5 \(\times\) cos(50\textdegree{}) \(\approx\) 0,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{CGM}\)) =  \(\dfrac{CM}{GM}\)
\par\addvspace{.5em}
 d'où sin(75\textdegree{}) = \(\dfrac{CM}{4{,}8}\)

On a donc CM = 4,8 \(\times\) sin(75\textdegree{}) \(\approx\) 4,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{DVT}\)) =  \(\dfrac{DT}{VT}\)
\par\addvspace{.5em}
 d'où sin(60\textdegree{}) = \(\dfrac{0{,}9}{VT}\)
\par\addvspace{.5em}
 On a donc VT = \(\dfrac{0{,}9}{\sin(60^{\circ})}\) \(\approx\) 1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.293,1.162) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{HKP}\)) = \(\dfrac{HP}{HK}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{HKP}\)) =  \(\dfrac{1{,}7}{4{,}7}\)
\par\addvspace{.5em}
 On a donc \(\widehat{HKP}\) =  ArcTan(\(\dfrac{1{,}7}{4{,}7}\)) \(\approx\) 20\textdegree{}\end{document}
