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\begin{document}

\sloppy

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

 Dans le triangle LZB rectangle en L, on sait que :
\begin{itemize}
\item LZ=2,1 cm
\item ZB=9,6 cm
\end{itemize}

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

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

 Dans le triangle DFG rectangle en D, on sait que :
\begin{itemize}
\item DF=8,9 cm
\item \(\widehat{FGD}\) =  36\textdegree{}
\end{itemize}

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

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

 Dans le triangle BAN rectangle en B, on sait que :
\begin{itemize}
\item BA=5,4 cm
\item \(\widehat{ANB}\) =  21\textdegree{}
\end{itemize}

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

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

 Dans le triangle CND rectangle en C, on sait que :
\begin{itemize}
\item CN=2,3 cm
\item CD=3,6 cm
\end{itemize}

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

 Dans le triangle VGH rectangle en V, on sait que :
\begin{itemize}
\item VH=4,1 cm
\item \(\widehat{VGH}\) = 56\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{LBZ}\)) = \(\dfrac{LZ}{ZB}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{LBZ}\)) =\(\dfrac{2{,}1}{9{,}6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{LBZ}\) = ArcSin(\(\dfrac{2{,}1}{9{,}6}\)) \(\approx\) 13\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{DGF}\)) = \(\dfrac{DF}{FG}\)
\par\addvspace{.5em}
 D'où sin(36\textdegree{}) = \(\dfrac{8{,}9}{FG}\)
\par\addvspace{.5em}
On a donc FG =\(\dfrac{8{,}9}{\sin(36^{\circ})}\) \(\approx\) 15,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{BNA}\)) = \(\dfrac{BA}{BN}\)
\par\addvspace{.5em}
d'où  tan(21\textdegree{}) =  \(\dfrac{5{,}4}{BN}\)
\par\addvspace{.5em}
On a donc BN = \(\dfrac{5{,}4}{\tan(21^{\circ})}\) \(\approx\) 14,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{CND}\)) = \(\dfrac{CD}{CN}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{CND}\)) = \(\dfrac{3{,}6}{2{,}3}\)
\par\addvspace{.5em}
 On a donc CND = ArcTan(\(\dfrac{3{,}6}{2{,}3}\)) \(\approx\) 57\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.924,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {4,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.737,2.745) {56°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{VGH}\)) = \(\dfrac{VH}{VG}\)
\par\addvspace{.5em}
tan(56\textdegree{}) = \(\dfrac{4{,}1}{VG}\)
\par\addvspace{.5em}
 On a donc VG = \(\dfrac{4{,}1}{\tan(56^{\circ})}\) \(\approx\) 2,8 cm\end{document}
