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\begin{document}

\sloppy

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

 Dans le triangle FWD rectangle en F, on sait que :
\begin{itemize}
\item FD=1,8 cm
\item \(\widehat{FWD}\) = 55\textdegree{}
\end{itemize}

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

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

 Dans le triangle DFV rectangle en D, on sait que :
\begin{itemize}
\item DF=2,7 cm
\item DV=6,2 cm
\end{itemize}

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

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

 Dans le triangle DGW rectangle en D, on sait que :
\begin{itemize}
\item DW=8,6 cm
\item \(\widehat{DGW}\) = 56\textdegree{}
\end{itemize}

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

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

 Dans le triangle KRS rectangle en K, on sait que :
\begin{itemize}
\item KR=1,7 cm
\item RS=7,8 cm
\end{itemize}

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

 Dans le triangle LWS rectangle en L, on sait que :
\begin{itemize}
\item LW=8 cm
\item \(\widehat{WSL}\) =  42\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{FWD}\)) = \(\dfrac{FD}{FW}\)
\par\addvspace{.5em}
tan(55\textdegree{}) = \(\dfrac{1{,}8}{FW}\)
\par\addvspace{.5em}
 On a donc FW = \(\dfrac{1{,}8}{\tan(55^{\circ})}\) \(\approx\) 1,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{DVF}\)) = \(\dfrac{DF}{DV}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{DVF}\)) =  \(\dfrac{2{,}7}{6{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{DVF}\) =  ArcTan(\(\dfrac{2{,}7}{6{,}2}\)) \(\approx\) 24\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{DGW}\)) =  \(\dfrac{DW}{GW}\)
\par\addvspace{.5em}
 d'où sin(56\textdegree{}) = \(\dfrac{8{,}6}{GW}\)
\par\addvspace{.5em}
 On a donc GW = \(\dfrac{8{,}6}{\sin(56^{\circ})}\) \(\approx\) 10,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{KRS}\)) = \(\dfrac{KR}{RS}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{KRS}\)) = \(\dfrac{1{,}7}{7{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{KRS}\) =ArcCos(\(\dfrac{1{,}7}{7{,}8}\)) \(\approx\)77\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.432,0.772) {42°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{LSW}\)) = \(\dfrac{LW}{LS}\)
\par\addvspace{.5em}
d'où  tan(42\textdegree{}) =  \(\dfrac{8}{LS}\)
\par\addvspace{.5em}
On a donc LS = \(\dfrac{8}{\tan(42^{\circ})}\) \(\approx\) 8,9 cm\end{document}
