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\begin{document}

\sloppy

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

 Dans le triangle PVF rectangle en P, on sait que :
\begin{itemize}
\item VF=4,1 cm
\item \(\widehat{VFP}\) =  19\textdegree{}
\end{itemize}

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

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

 Dans le triangle GCF rectangle en G, on sait que :
\begin{itemize}
\item CF=3,9 cm
\item \(\widehat{GCF}\) = 49\textdegree{}
\end{itemize}

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

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

 Dans le triangle TDL rectangle en T, on sait que :
\begin{itemize}
\item TD=3,1 cm
\item DL=7 cm
\end{itemize}

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

 Dans le triangle SMZ rectangle en S, on sait que :
\begin{itemize}
\item SZ=6,6 cm
\item MZ=8,2 cm
\end{itemize}

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

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

 Dans le triangle BTJ rectangle en B, on sait que :
\begin{itemize}
\item BT=1,7 cm
\item \(\widehat{TJB}\) =  38\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{PFV}\)) = \(\dfrac{PV}{VF}\)
\par\addvspace{.5em}
 D'où sin(19\textdegree{}) = \(\dfrac{PV}{4{,}1}\)

On a donc PV = 4,1 \(\times\) sin(19\textdegree{}) \(\approx\) 1,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{GCF}\)) =  \(\dfrac{GF}{CF}\)
\par\addvspace{.5em}
 d'où sin(49\textdegree{}) = \(\dfrac{GF}{3{,}9}\)

On a donc GF = 3,9 \(\times\) sin(49\textdegree{}) \(\approx\) 2,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{TDL}\)) = \(\dfrac{TD}{DL}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{TDL}\)) = \(\dfrac{3{,}1}{7}\)
\par\addvspace{.5em}
On a donc \(\widehat{TDL}\) =ArcCos(\(\dfrac{3{,}1}{7}\)) \(\approx\)64\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{SZM}\)) = \(\dfrac{SZ}{MZ}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{SZM}\)) =  \(\dfrac{6{,}6}{8{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{SZM}\) = Arccos (\(\dfrac{6{,}6}{8{,}2}\)) \(\approx\) 36\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{BJT}\)) = \(\dfrac{BT}{TJ}\)
\par\addvspace{.5em}
 D'où sin(38\textdegree{}) = \(\dfrac{1{,}7}{TJ}\)
\par\addvspace{.5em}
On a donc TJ =\(\dfrac{1{,}7}{\sin(38^{\circ})}\) \(\approx\) 2,8 cm\end{document}
