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\begin{document}

\sloppy

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

 Dans le triangle FVH rectangle en F, on sait que :
\begin{itemize}
\item FH=5,8 cm
\item VH=8,5 cm
\end{itemize}

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

 Dans le triangle SDR rectangle en S, on sait que :
\begin{itemize}
\item DR=1,4 cm
\item \(\widehat{DRS}\) =  21\textdegree{}
\end{itemize}

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

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

 Dans le triangle NRA rectangle en N, on sait que :
\begin{itemize}
\item NA=6,8 cm
\item \(\widehat{NRA}\) = 48\textdegree{}
\end{itemize}

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

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

 Dans le triangle TZG rectangle en T, on sait que :
\begin{itemize}
\item TZ=6,7 cm
\item \(\widehat{TZG}\) = 77\textdegree{}
\end{itemize}

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

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

 Dans le triangle CTA rectangle en C, on sait que :
\begin{itemize}
\item CT=2 cm
\item TA=10 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{FVH}\)) =  \(\dfrac{FH}{VH}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{FVH}\)) = \(\dfrac{5{,}8}{8{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{FVH}\) =  ArcSin(\(\dfrac{5{,}8}{8{,}5}\)) \(\approx\) 43\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{SRD}\)) = \(\dfrac{SD}{DR}\)
\par\addvspace{.5em}
 D'où sin(21\textdegree{}) = \(\dfrac{SD}{1{,}4}\)

On a donc SD = 1,4 \(\times\) sin(21\textdegree{}) \(\approx\) 0,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{NRA}\)) =  \(\dfrac{NA}{RA}\)
\par\addvspace{.5em}
 d'où sin(48\textdegree{}) = \(\dfrac{6{,}8}{RA}\)
\par\addvspace{.5em}
 On a donc RA = \(\dfrac{6{,}8}{\sin(48^{\circ})}\) \(\approx\) 9,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{TZG}\)) = \(\dfrac{TG}{TZ}\)
\par\addvspace{.5em}
d'où tan(77\textdegree{}) =\(\dfrac{TG}{6{,}7}\)
\par\addvspace{.5em}
On a donc TG = 6,7 \(\times\) tan(77\textdegree{}) \(\approx\) 29 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{CAT}\)) = \(\dfrac{CT}{TA}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{CAT}\)) =\(\dfrac{2}{10}\)
\par\addvspace{.5em}
 On a donc \(\widehat{CAT}\) = ArcSin(\(\dfrac{2}{10}\)) \(\approx\) 12\textdegree{}\end{document}
