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\begin{document}

\sloppy

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

 Dans le triangle ZGS rectangle en Z, on sait que :
\begin{itemize}
\item ZG=1,1 cm
\item GS=8 cm
\end{itemize}

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

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

 Dans le triangle FBG rectangle en F, on sait que :
\begin{itemize}
\item FG=9,7 cm
\item \(\widehat{FBG}\) = 52\textdegree{}
\end{itemize}

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

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

 Dans le triangle BLR rectangle en B, on sait que :
\begin{itemize}
\item BL=2,6 cm
\item \(\widehat{LRB}\) =  43\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZNP rectangle en Z, on sait que :
\begin{itemize}
\item ZP=4,5 cm
\item NP=9,6 cm
\end{itemize}

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

 Dans le triangle LCB rectangle en L, on sait que :
\begin{itemize}
\item LB=7,6 cm
\item \(\widehat{CBL}\) =  17\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{ZSG}\)) = \(\dfrac{ZG}{GS}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{ZSG}\)) =\(\dfrac{1{,}1}{8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ZSG}\) = ArcSin(\(\dfrac{1{,}1}{8}\)) \(\approx\) 8\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{FBG}\)) = \(\dfrac{FG}{FB}\)
\par\addvspace{.5em}
tan(52\textdegree{}) = \(\dfrac{9{,}7}{FB}\)
\par\addvspace{.5em}
 On a donc FB = \(\dfrac{9{,}7}{\tan(52^{\circ})}\) \(\approx\) 7,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{BRL}\)) = \(\dfrac{BL}{BR}\)
\par\addvspace{.5em}
d'où  tan(43\textdegree{}) =  \(\dfrac{2{,}6}{BR}\)
\par\addvspace{.5em}
On a donc BR = \(\dfrac{2{,}6}{\tan(43^{\circ})}\) \(\approx\) 2,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{ZNP}\)) =  \(\dfrac{ZP}{NP}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{ZNP}\)) = \(\dfrac{4{,}5}{9{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{ZNP}\) =  ArcSin(\(\dfrac{4{,}5}{9{,}6}\)) \(\approx\) 28\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.291,1.283) {17°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{LBC}\)) = \(\dfrac{LB}{CB}\)
\par\addvspace{.5em}
  d'où    cos(17\textdegree{}) =  \(\dfrac{7{,}6}{CB}\)
\par\addvspace{.5em}
 On a donc CB =  \(\dfrac{7{,}6}{\cos(17^{\circ})}\) \(\approx\) 7,9 cm\end{document}
