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\begin{document}

\sloppy

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

 Dans le triangle WZF rectangle en W, on sait que :
\begin{itemize}
\item WF=4,6 cm
\item ZF=7,8 cm
\end{itemize}

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

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

 Dans le triangle WDT rectangle en W, on sait que :
\begin{itemize}
\item DT=2,1 cm
\item \(\widehat{WDT}\) = 79\textdegree{}
\end{itemize}

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

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

 Dans le triangle LNF rectangle en L, on sait que :
\begin{itemize}
\item NF=1,3 cm
\item \(\widehat{LNF}\) = 77\textdegree{}
\end{itemize}

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

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

 Dans le triangle HVW rectangle en H, on sait que :
\begin{itemize}
\item HV=2 cm
\item HW=6 cm
\end{itemize}

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

 Dans le triangle LGA rectangle en L, on sait que :
\begin{itemize}
\item LA=4,8 cm
\item \(\widehat{LGA}\) = 50\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{WFZ}\)) = \(\dfrac{WF}{ZF}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{WFZ}\)) =  \(\dfrac{4{,}6}{7{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{WFZ}\) = Arccos (\(\dfrac{4{,}6}{7{,}8}\)) \(\approx\) 54\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{WDT}\)) =  \(\dfrac{WT}{DT}\)
\par\addvspace{.5em}
 d'où sin(79\textdegree{}) = \(\dfrac{WT}{2{,}1}\)

On a donc WT = 2,1 \(\times\) sin(79\textdegree{}) \(\approx\) 2,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{LNF}\)) = \(\dfrac{LN}{NF}\)
\par\addvspace{.5em}
d'où cos(77\textdegree{}) = \(\dfrac{LN}{1{,}3}\)

On a donc LN = 1,3 \(\times\) cos(77\textdegree{}) \(\approx\) 0,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{HVW}\)) = \(\dfrac{HW}{HV}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{HVW}\)) = \(\dfrac{6}{2}\)
\par\addvspace{.5em}
 On a donc HVW = ArcTan(\(\dfrac{6}{2}\)) \(\approx\) 72\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{LGA}\)) =  \(\dfrac{LA}{GA}\)
\par\addvspace{.5em}
 d'où sin(50\textdegree{}) = \(\dfrac{4{,}8}{GA}\)
\par\addvspace{.5em}
 On a donc GA = \(\dfrac{4{,}8}{\sin(50^{\circ})}\) \(\approx\) 6,3 cm\end{document}
