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\begin{document}

\sloppy

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

 Dans le triangle DAF rectangle en D, on sait que :
\begin{itemize}
\item DA=1,2 cm
\item \(\widehat{AFD}\) =  13\textdegree{}
\end{itemize}

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

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

 Dans le triangle CWL rectangle en C, on sait que :
\begin{itemize}
\item CW=3,2 cm
\item CL=4,8 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{CWL}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle MZP rectangle en M, on sait que :
\begin{itemize}
\item MZ=1 cm
\item \(\widehat{MZP}\) = 65\textdegree{}
\end{itemize}

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

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

 Dans le triangle KDC rectangle en K, on sait que :
\begin{itemize}
\item KD=1 cm
\item DC=8 cm
\end{itemize}

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

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

 Dans le triangle LZH rectangle en L, on sait que :
\begin{itemize}
\item ZH=3,5 cm
\item \(\widehat{ZHL}\) =  24\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{DFA}\)) = \(\dfrac{DA}{DF}\)
\par\addvspace{.5em}
d'où  tan(13\textdegree{}) =  \(\dfrac{1{,}2}{DF}\)
\par\addvspace{.5em}
On a donc DF = \(\dfrac{1{,}2}{\tan(13^{\circ})}\) \(\approx\) 5,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{CWL}\)) = \(\dfrac{CL}{CW}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{CWL}\)) = \(\dfrac{4{,}8}{3{,}2}\)
\par\addvspace{.5em}
 On a donc CWL = ArcTan(\(\dfrac{4{,}8}{3{,}2}\)) \(\approx\) 56\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{MZP}\)) = \(\dfrac{MZ}{ZP}\)
\par\addvspace{.5em}
 d'où cos(65\textdegree{}) = \(\dfrac{1}{ZP}\)
\par\addvspace{.5em}
On a donc ZP = \(\dfrac{1}{\cos(65^{\circ})}\)\(\approx\) 2,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{KCD}\)) = \(\dfrac{KD}{DC}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{KCD}\)) =\(\dfrac{1}{8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{KCD}\) = ArcSin(\(\dfrac{1}{8}\)) \(\approx\) 7\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {3,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.298,0.981) {24°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{LHZ}\)) = \(\dfrac{LZ}{ZH}\)
\par\addvspace{.5em}
 D'où sin(24\textdegree{}) = \(\dfrac{LZ}{3{,}5}\)

On a donc LZ = 3,5 \(\times\) sin(24\textdegree{}) \(\approx\) 1,4 cm\end{document}
