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\begin{document}

\sloppy

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

 Dans le triangle DHG rectangle en D, on sait que :
\begin{itemize}
\item DG=5 cm
\item HG=9,3 cm
\end{itemize}

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

 Dans le triangle CKF rectangle en C, on sait que :
\begin{itemize}
\item CF=5,1 cm
\item \(\widehat{CKF}\) = 65\textdegree{}
\end{itemize}

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

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

 Dans le triangle LWH rectangle en L, on sait que :
\begin{itemize}
\item LW=4,1 cm
\item \(\widehat{LWH}\) = 65\textdegree{}
\end{itemize}

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

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

 Dans le triangle DZV rectangle en D, on sait que :
\begin{itemize}
\item DZ=2,2 cm
\item DV=5,5 cm
\end{itemize}

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

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

 Dans le triangle DJL rectangle en D, on sait que :
\begin{itemize}
\item DJ=5,7 cm
\item \(\widehat{DJL}\) = 69\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [LJ]. (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 (2.794,2.693) {?};
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 Dans le triangle DHG rectangle en D, on cherche une relation entre l'angle aigu   \(\widehat{DHG}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{DHG}\)) =  \(\dfrac{DG}{HG}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{DHG}\)) = \(\dfrac{5}{9{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{DHG}\) =  ArcSin(\(\dfrac{5}{9{,}3}\)) \(\approx\) 33\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{CKF}\)) = \(\dfrac{CF}{CK}\)
\par\addvspace{.5em}
tan(65\textdegree{}) = \(\dfrac{5{,}1}{CK}\)
\par\addvspace{.5em}
 On a donc CK = \(\dfrac{5{,}1}{\tan(65^{\circ})}\) \(\approx\) 2,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{LWH}\)) = \(\dfrac{LH}{LW}\)
\par\addvspace{.5em}
d'où tan(65\textdegree{}) =\(\dfrac{LH}{4{,}1}\)
\par\addvspace{.5em}
On a donc LH = 4,1 \(\times\) tan(65\textdegree{}) \(\approx\) 8,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{DVZ}\)) = \(\dfrac{DZ}{DV}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{DVZ}\)) =  \(\dfrac{2{,}2}{5{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{DVZ}\) =  ArcTan(\(\dfrac{2{,}2}{5{,}5}\)) \(\approx\) 22\textdegree{}
\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 (1.428,2.324) {69°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{DJL}\)) = \(\dfrac{DJ}{JL}\)
\par\addvspace{.5em}
 d'où cos(69\textdegree{}) = \(\dfrac{5{,}7}{JL}\)
\par\addvspace{.5em}
On a donc JL = \(\dfrac{5{,}7}{\cos(69^{\circ})}\)\(\approx\) 15,9 cm\end{document}
