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\begin{document}

\sloppy

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

 Dans le triangle BCN rectangle en B, on sait que :
\begin{itemize}
\item BN=5,7 cm
\item CN=7,6 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{BCN}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle WBJ rectangle en W, on sait que :
\begin{itemize}
\item WB=2,6 cm
\item BJ=9,8 cm
\end{itemize}

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

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

 Dans le triangle HGC rectangle en H, on sait que :
\begin{itemize}
\item HC=2,1 cm
\item \(\widehat{GCH}\) =  23\textdegree{}
\end{itemize}

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

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

 Dans le triangle KRL rectangle en K, on sait que :
\begin{itemize}
\item KR=1 cm
\item \(\widehat{KRL}\) = 67\textdegree{}
\end{itemize}

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

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

 Dans le triangle PAC rectangle en P, on sait que :
\begin{itemize}
\item PA=4,9 cm
\item \(\widehat{ACP}\) =  39\textdegree{}
\end{itemize}

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

\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 (2.187,2.726) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{BCN}\)) =  \(\dfrac{BN}{CN}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{BCN}\)) = \(\dfrac{5{,}7}{7{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{BCN}\) =  ArcSin(\(\dfrac{5{,}7}{7{,}6}\)) \(\approx\) 49\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{WJB}\)) = \(\dfrac{WB}{BJ}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{WJB}\)) =\(\dfrac{2{,}6}{9{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{WJB}\) = ArcSin(\(\dfrac{2{,}6}{9{,}8}\)) \(\approx\) 15\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{HCG}\)) = \(\dfrac{HG}{HC}\)
\par\addvspace{.5em}
d'où  tan(23\textdegree{}) =  \(\dfrac{HG}{2{,}1}\)
\par\addvspace{.5em}
On a donc HG =  2,1 \(\times\) tan(23\textdegree{})  \(\approx\) 0,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{KRL}\)) = \(\dfrac{KR}{RL}\)
\par\addvspace{.5em}
 d'où cos(67\textdegree{}) = \(\dfrac{1}{RL}\)
\par\addvspace{.5em}
On a donc RL = \(\dfrac{1}{\cos(67^{\circ})}\)\(\approx\) 2,6 cm
\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 (4.600,0.756) {39°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{PCA}\)) = \(\dfrac{PA}{PC}\)
\par\addvspace{.5em}
d'où  tan(39\textdegree{}) =  \(\dfrac{4{,}9}{PC}\)
\par\addvspace{.5em}
On a donc PC = \(\dfrac{4{,}9}{\tan(39^{\circ})}\) \(\approx\) 6,1 cm\end{document}
