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\begin{document}

\sloppy

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

 Dans le triangle ZBT rectangle en Z, on sait que :
\begin{itemize}
\item BT=0,5 cm
\item \(\widehat{BTZ}\) =  23\textdegree{}
\end{itemize}

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

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

 Dans le triangle JDF rectangle en J, on sait que :
\begin{itemize}
\item DF=3,2 cm
\item \(\widehat{DFJ}\) =  42\textdegree{}
\end{itemize}

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

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

 Dans le triangle TGV rectangle en T, on sait que :
\begin{itemize}
\item TV=8 cm
\item \(\widehat{TGV}\) = 48\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZTR rectangle en Z, on sait que :
\begin{itemize}
\item ZR=4,4 cm
\item TR=7 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{ZTR}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle WCA rectangle en W, on sait que :
\begin{itemize}
\item WC=1,1 cm
\item WA=3,6 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{ZTB}\)) = \(\dfrac{ZB}{BT}\)
\par\addvspace{.5em}
 D'où sin(23\textdegree{}) = \(\dfrac{ZB}{0{,}5}\)

On a donc ZB = 0,5 \(\times\) sin(23\textdegree{}) \(\approx\) 0,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{JFD}\)) = \(\dfrac{JF}{DF}\)
\par\addvspace{.5em}
  d'où    cos(42\textdegree{}) =  \(\dfrac{JF}{3{,}2}\)

On a donc JF = 3,2 \(\times\) cos(42\textdegree{}) \(\approx\) 2{,}4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{TGV}\)) =  \(\dfrac{TV}{GV}\)
\par\addvspace{.5em}
 d'où sin(48\textdegree{}) = \(\dfrac{8}{GV}\)
\par\addvspace{.5em}
 On a donc GV = \(\dfrac{8}{\sin(48^{\circ})}\) \(\approx\) 10,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{ZTR}\)) =  \(\dfrac{ZR}{TR}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{ZTR}\)) = \(\dfrac{4{,}4}{7}\)
\par\addvspace{.5em}
On a donc \(\widehat{ZTR}\) =  ArcSin(\(\dfrac{4{,}4}{7}\)) \(\approx\) 39\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
tan(\(\widehat{WAC}\)) = \(\dfrac{WC}{WA}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{WAC}\)) =  \(\dfrac{1{,}1}{3{,}6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{WAC}\) =  ArcTan(\(\dfrac{1{,}1}{3{,}6}\)) \(\approx\) 17\textdegree{}\end{document}
