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\begin{document}

\sloppy

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

 Dans le triangle GHB rectangle en G, on sait que :
\begin{itemize}
\item GB=3,8 cm
\item HB=8,9 cm
\end{itemize}

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

 Dans le triangle CWB rectangle en C, on sait que :
\begin{itemize}
\item CB=6,3 cm
\item \(\widehat{WBC}\) =  31\textdegree{}
\end{itemize}

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

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

 Dans le triangle CJK rectangle en C, on sait que :
\begin{itemize}
\item CJ=1,4 cm
\item JK=8,8 cm
\end{itemize}

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

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

 Dans le triangle PDA rectangle en P, on sait que :
\begin{itemize}
\item PD=9 cm
\item \(\widehat{DAP}\) =  22\textdegree{}
\end{itemize}

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

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

 Dans le triangle VHN rectangle en V, on sait que :
\begin{itemize}
\item VH=3,5 cm
\item \(\widehat{VHN}\) = 58\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [NH]. (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 (2.985,2.683) {?};
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 Dans le triangle GHB rectangle en G, on cherche une relation entre l'angle aigu   \(\widehat{GHB}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{GHB}\)) =  \(\dfrac{GB}{HB}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{GHB}\)) = \(\dfrac{3{,}8}{8{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{GHB}\) =  ArcSin(\(\dfrac{3{,}8}{8{,}9}\)) \(\approx\) 25\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{CBW}\)) = \(\dfrac{CW}{CB}\)
\par\addvspace{.5em}
d'où  tan(31\textdegree{}) =  \(\dfrac{CW}{6{,}3}\)
\par\addvspace{.5em}
On a donc CW =  6,3 \(\times\) tan(31\textdegree{})  \(\approx\) 3,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{CKJ}\)) = \(\dfrac{CJ}{JK}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{CKJ}\)) =\(\dfrac{1{,}4}{8{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{CKJ}\) = ArcSin(\(\dfrac{1{,}4}{8{,}8}\)) \(\approx\) 9\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
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 Dans le triangle PDA rectangle en P, on cherche une relation entre l'angle aigu   \(\widehat{PAD}\) son coté opposé et son coté adjacent.

\par\addvspace{.5em}
tan(\(\widehat{PAD}\)) = \(\dfrac{PD}{PA}\)
\par\addvspace{.5em}
d'où  tan(22\textdegree{}) =  \(\dfrac{9}{PA}\)
\par\addvspace{.5em}
On a donc PA = \(\dfrac{9}{\tan(22^{\circ})}\) \(\approx\) 22,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.583,2.750) {58°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{VHN}\)) = \(\dfrac{VH}{HN}\)
\par\addvspace{.5em}
 d'où cos(58\textdegree{}) = \(\dfrac{3{,}5}{HN}\)
\par\addvspace{.5em}
On a donc HN = \(\dfrac{3{,}5}{\cos(58^{\circ})}\)\(\approx\) 6,6 cm\end{document}
