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\begin{document}

\sloppy

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

 Dans le triangle LFC rectangle en L, on sait que :
\begin{itemize}
\item LF=1,5 cm
\item FC=7,2 cm
\end{itemize}

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

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

 Dans le triangle DWA rectangle en D, on sait que :
\begin{itemize}
\item DA=9,2 cm
\item \(\widehat{DWA}\) = 71\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 GFB rectangle en G, on sait que :
\begin{itemize}
\item GF=5,8 cm
\item \(\widehat{FBG}\) =  30\textdegree{}
\end{itemize}

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

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

 Dans le triangle BKG rectangle en B, on sait que :
\begin{itemize}
\item BG=7,6 cm
\item \(\widehat{KGB}\) =  14\textdegree{}
\end{itemize}

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

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

 Dans le triangle LSG rectangle en L, on sait que :
\begin{itemize}
\item LS=2,5 cm
\item LG=5,4 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{LSG}\). \newpage
\section*{\textcolor{red}{Correction}}

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

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.287,1.481) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{LCF}\)) = \(\dfrac{LF}{FC}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{LCF}\)) =\(\dfrac{1{,}5}{7{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{LCF}\) = ArcSin(\(\dfrac{1{,}5}{7{,}2}\)) \(\approx\) 12\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{DWA}\)) =  \(\dfrac{DA}{WA}\)
\par\addvspace{.5em}
 d'où sin(71\textdegree{}) = \(\dfrac{9{,}2}{WA}\)
\par\addvspace{.5em}
 On a donc WA = \(\dfrac{9{,}2}{\sin(71^{\circ})}\) \(\approx\) 9,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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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 (5.281,0.706) {30°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{GBF}\)) = \(\dfrac{GF}{GB}\)
\par\addvspace{.5em}
d'où  tan(30\textdegree{}) =  \(\dfrac{5{,}8}{GB}\)
\par\addvspace{.5em}
On a donc GB = \(\dfrac{5{,}8}{\tan(30^{\circ})}\) \(\approx\) 10 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.288,1.404) {14°};
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 Dans le triangle BKG rectangle en B, on cherche une relation entre l'angle aigu   \(\widehat{BGK}\) son coté opposé et son coté adjacent.

\par\addvspace{.5em}
tan(\(\widehat{BGK}\)) = \(\dfrac{BK}{BG}\)
\par\addvspace{.5em}
d'où  tan(14\textdegree{}) =  \(\dfrac{BK}{7{,}6}\)
\par\addvspace{.5em}
On a donc BK =  7,6 \(\times\) tan(14\textdegree{})  \(\approx\) 1,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.201) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.732) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.532) {5,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.409,2.506) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{LSG}\)) = \(\dfrac{LG}{LS}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{LSG}\)) = \(\dfrac{5{,}4}{2{,}5}\)
\par\addvspace{.5em}
 On a donc LSG = ArcTan(\(\dfrac{5{,}4}{2{,}5}\)) \(\approx\) 65\textdegree{}\end{document}
