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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 DF=3 cm
\item \(\widehat{AFD}\) =  17\textdegree{}
\end{itemize}

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

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

 Dans le triangle JNS rectangle en J, on sait que :
\begin{itemize}
\item JN=1,4 cm
\item NS=9,3 cm
\end{itemize}

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

 Dans le triangle KFG rectangle en K, on sait que :
\begin{itemize}
\item KG=4,4 cm
\item \(\widehat{FGK}\) =  22\textdegree{}
\end{itemize}

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

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

 Dans le triangle DHL rectangle en D, on sait que :
\begin{itemize}
\item DH=1,8 cm
\item DL=4,4 cm
\end{itemize}

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

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

 Dans le triangle MKZ rectangle en M, on sait que :
\begin{itemize}
\item KZ=9,4 cm
\item \(\widehat{KZM}\) =  43\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [MZ]. (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(17\textdegree{}) =  \(\dfrac{DA}{3}\)
\par\addvspace{.5em}
On a donc DA =  3 \(\times\) tan(17\textdegree{})  \(\approx\) 0,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{JNS}\)) = \(\dfrac{JN}{NS}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{JNS}\)) = \(\dfrac{1{,}4}{9{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{JNS}\) =ArcCos(\(\dfrac{1{,}4}{9{,}3}\)) \(\approx\)81\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{KGF}\)) = \(\dfrac{KG}{FG}\)
\par\addvspace{.5em}
  d'où    cos(22\textdegree{}) =  \(\dfrac{4{,}4}{FG}\)
\par\addvspace{.5em}
 On a donc FG =  \(\dfrac{4{,}4}{\cos(22^{\circ})}\) \(\approx\) 4,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{DLH}\)) = \(\dfrac{DH}{DL}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{DLH}\)) =  \(\dfrac{1{,}8}{4{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{DLH}\) =  ArcTan(\(\dfrac{1{,}8}{4{,}4}\)) \(\approx\) 22\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.381,0.778) {43°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{MZK}\)) = \(\dfrac{MZ}{KZ}\)
\par\addvspace{.5em}
  d'où    cos(43\textdegree{}) =  \(\dfrac{MZ}{9{,}4}\)

On a donc MZ = 9,4 \(\times\) cos(43\textdegree{}) \(\approx\) 6{,}9 cm\end{document}
