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\begin{document}

\sloppy

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

 Dans le triangle DAV rectangle en D, on sait que :
\begin{itemize}
\item AV=9,1 cm
\item \(\widehat{DAV}\) = 79\textdegree{}
\end{itemize}

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

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

 Dans le triangle PRL rectangle en P, on sait que :
\begin{itemize}
\item RL=1,4 cm
\item \(\widehat{PRL}\) = 55\textdegree{}
\end{itemize}

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

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

 Dans le triangle DGL rectangle en D, on sait que :
\begin{itemize}
\item DL=5,9 cm
\item GL=9 cm
\end{itemize}

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

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

 Dans le triangle WCB rectangle en W, on sait que :
\begin{itemize}
\item WC=2,2 cm
\item WB=3,8 cm
\end{itemize}

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

 Dans le triangle LTA rectangle en L, on sait que :
\begin{itemize}
\item LA=9,5 cm
\item \(\widehat{LTA}\) = 49\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{DAV}\)) =  \(\dfrac{DV}{AV}\)
\par\addvspace{.5em}
 d'où sin(79\textdegree{}) = \(\dfrac{DV}{9{,}1}\)

On a donc DV = 9,1 \(\times\) sin(79\textdegree{}) \(\approx\) 8,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{PRL}\)) = \(\dfrac{PR}{RL}\)
\par\addvspace{.5em}
d'où cos(55\textdegree{}) = \(\dfrac{PR}{1{,}4}\)

On a donc PR = 1,4 \(\times\) cos(55\textdegree{}) \(\approx\) 0,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{DLG}\)) = \(\dfrac{DL}{GL}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{DLG}\)) =  \(\dfrac{5{,}9}{9}\)
\par\addvspace{.5em}
On a donc \(\widehat{DLG}\) = Arccos (\(\dfrac{5{,}9}{9}\)) \(\approx\) 49\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{WCB}\)) = \(\dfrac{WB}{WC}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{WCB}\)) = \(\dfrac{3{,}8}{2{,}2}\)
\par\addvspace{.5em}
 On a donc WCB = ArcTan(\(\dfrac{3{,}8}{2{,}2}\)) \(\approx\) 60\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{LTA}\)) =  \(\dfrac{LA}{TA}\)
\par\addvspace{.5em}
 d'où sin(49\textdegree{}) = \(\dfrac{9{,}5}{TA}\)
\par\addvspace{.5em}
 On a donc TA = \(\dfrac{9{,}5}{\sin(49^{\circ})}\) \(\approx\) 12,6 cm\end{document}
