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\begin{document}

\sloppy

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

 Dans le triangle PVW rectangle en P, on sait que :
\begin{itemize}
\item PW=3,6 cm
\item VW=9,4 cm
\end{itemize}

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

 Dans le triangle AWF rectangle en A, on sait que :
\begin{itemize}
\item WF=8,7 cm
\item \(\widehat{AWF}\) = 50\textdegree{}
\end{itemize}

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

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

 Dans le triangle GWV rectangle en G, on sait que :
\begin{itemize}
\item GV=5,7 cm
\item WV=7,1 cm
\end{itemize}

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

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

 Dans le triangle VWZ rectangle en V, on sait que :
\begin{itemize}
\item VZ=9,3 cm
\item \(\widehat{VWZ}\) = 68\textdegree{}
\end{itemize}

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

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

 Dans le triangle LHT rectangle en L, on sait que :
\begin{itemize}
\item HT=7,8 cm
\item \(\widehat{LHT}\) = 47\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{PVW}\)) =  \(\dfrac{PW}{VW}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{PVW}\)) = \(\dfrac{3{,}6}{9{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{PVW}\) =  ArcSin(\(\dfrac{3{,}6}{9{,}4}\)) \(\approx\) 23\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{AWF}\)) = \(\dfrac{AW}{WF}\)
\par\addvspace{.5em}
d'où cos(50\textdegree{}) = \(\dfrac{AW}{8{,}7}\)

On a donc AW = 8,7 \(\times\) cos(50\textdegree{}) \(\approx\) 5,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{GVW}\)) = \(\dfrac{GV}{WV}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{GVW}\)) =  \(\dfrac{5{,}7}{7{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{GVW}\) = Arccos (\(\dfrac{5{,}7}{7{,}1}\)) \(\approx\) 37\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{VWZ}\)) =  \(\dfrac{VZ}{WZ}\)
\par\addvspace{.5em}
 d'où sin(68\textdegree{}) = \(\dfrac{9{,}3}{WZ}\)
\par\addvspace{.5em}
 On a donc WZ = \(\dfrac{9{,}3}{\sin(68^{\circ})}\) \(\approx\) 10 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{LHT}\)) =  \(\dfrac{LT}{HT}\)
\par\addvspace{.5em}
 d'où sin(47\textdegree{}) = \(\dfrac{LT}{7{,}8}\)

On a donc LT = 7,8 \(\times\) sin(47\textdegree{}) \(\approx\) 5,7 cm\end{document}
