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\begin{document}

\sloppy

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

 Dans le triangle PTH rectangle en P, on sait que :
\begin{itemize}
\item PH=6,5 cm
\item TH=9,8 cm
\end{itemize}

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

 Dans le triangle PAM rectangle en P, on sait que :
\begin{itemize}
\item PM=8,8 cm
\item \(\widehat{AMP}\) =  20\textdegree{}
\end{itemize}

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

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

 Dans le triangle FNS rectangle en F, on sait que :
\begin{itemize}
\item FS=2,6 cm
\item \(\widehat{NSF}\) =  36\textdegree{}
\end{itemize}

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

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

 Dans le triangle VZW rectangle en V, on sait que :
\begin{itemize}
\item VZ=2 cm
\item VW=4 cm
\end{itemize}

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

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

 Dans le triangle TRD rectangle en T, on sait que :
\begin{itemize}
\item TR=4,3 cm
\item \(\widehat{RDT}\) =  38\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{PTH}\)) =  \(\dfrac{PH}{TH}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{PTH}\)) = \(\dfrac{6{,}5}{9{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{PTH}\) =  ArcSin(\(\dfrac{6{,}5}{9{,}8}\)) \(\approx\) 42\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{PMA}\)) = \(\dfrac{PA}{PM}\)
\par\addvspace{.5em}
d'où  tan(20\textdegree{}) =  \(\dfrac{PA}{8{,}8}\)
\par\addvspace{.5em}
On a donc PA =  8,8 \(\times\) tan(20\textdegree{})  \(\approx\) 3,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{FSN}\)) = \(\dfrac{FS}{NS}\)
\par\addvspace{.5em}
  d'où    cos(36\textdegree{}) =  \(\dfrac{2{,}6}{NS}\)
\par\addvspace{.5em}
 On a donc NS =  \(\dfrac{2{,}6}{\cos(36^{\circ})}\) \(\approx\) 3,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{VWZ}\)) = \(\dfrac{VZ}{VW}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{VWZ}\)) =  \(\dfrac{2}{4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{VWZ}\) =  ArcTan(\(\dfrac{2}{4}\)) \(\approx\) 27\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.662,0.750) {38°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{TDR}\)) = \(\dfrac{TR}{TD}\)
\par\addvspace{.5em}
d'où  tan(38\textdegree{}) =  \(\dfrac{4{,}3}{TD}\)
\par\addvspace{.5em}
On a donc TD = \(\dfrac{4{,}3}{\tan(38^{\circ})}\) \(\approx\) 5,5 cm\end{document}
