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\begin{document}

\sloppy

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

 Dans le triangle NTL rectangle en N, on sait que :
\begin{itemize}
\item NT=2,5 cm
\item NL=4 cm
\end{itemize}

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

 Dans le triangle RSA rectangle en R, on sait que :
\begin{itemize}
\item SA=6 cm
\item \(\widehat{RSA}\) = 46\textdegree{}
\end{itemize}

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

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

 Dans le triangle AVH rectangle en A, on sait que :
\begin{itemize}
\item AV=2,5 cm
\item VH=7 cm
\end{itemize}

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

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

 Dans le triangle VHA rectangle en V, on sait que :
\begin{itemize}
\item HA=5,7 cm
\item \(\widehat{VHA}\) = 62\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZHT rectangle en Z, on sait que :
\begin{itemize}
\item ZH=8,8 cm
\item \(\widehat{HTZ}\) =  34\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{NTL}\)) = \(\dfrac{NL}{NT}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{NTL}\)) = \(\dfrac{4}{2{,}5}\)
\par\addvspace{.5em}
 On a donc NTL = ArcTan(\(\dfrac{4}{2{,}5}\)) \(\approx\) 58\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{RSA}\)) =  \(\dfrac{RA}{SA}\)
\par\addvspace{.5em}
 d'où sin(46\textdegree{}) = \(\dfrac{RA}{6}\)

On a donc RA = 6 \(\times\) sin(46\textdegree{}) \(\approx\) 4,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{AHV}\)) = \(\dfrac{AV}{VH}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{AHV}\)) =\(\dfrac{2{,}5}{7}\)
\par\addvspace{.5em}
 On a donc \(\widehat{AHV}\) = ArcSin(\(\dfrac{2{,}5}{7}\)) \(\approx\) 21\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{VHA}\)) = \(\dfrac{VH}{HA}\)
\par\addvspace{.5em}
d'où cos(62\textdegree{}) = \(\dfrac{VH}{5{,}7}\)

On a donc VH = 5,7 \(\times\) cos(62\textdegree{}) \(\approx\) 2,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{ZTH}\)) = \(\dfrac{ZH}{HT}\)
\par\addvspace{.5em}
 D'où sin(34\textdegree{}) = \(\dfrac{8{,}8}{HT}\)
\par\addvspace{.5em}
On a donc HT =\(\dfrac{8{,}8}{\sin(34^{\circ})}\) \(\approx\) 15,7 cm\end{document}
