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\begin{document}

\sloppy

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

 Dans le triangle DKR rectangle en D, on sait que :
\begin{itemize}
\item DR=5 cm
\item KR=8,4 cm
\end{itemize}

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

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

 Dans le triangle SZN rectangle en S, on sait que :
\begin{itemize}
\item SZ=1,6 cm
\item ZN=7,9 cm
\end{itemize}

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

 Dans le triangle GTN rectangle en G, on sait que :
\begin{itemize}
\item TN=9,7 cm
\item \(\widehat{TNG}\) =  40\textdegree{}
\end{itemize}

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

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

 Dans le triangle FDW rectangle en F, on sait que :
\begin{itemize}
\item FD=9,9 cm
\item \(\widehat{FDW}\) = 68\textdegree{}
\end{itemize}

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

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

 Dans le triangle LAG rectangle en L, on sait que :
\begin{itemize}
\item LG=5,9 cm
\item \(\widehat{AGL}\) =  28\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [GA]. (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 (3.942,0.833) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{DRK}\)) = \(\dfrac{DR}{KR}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{DRK}\)) =  \(\dfrac{5}{8{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{DRK}\) = Arccos (\(\dfrac{5}{8{,}4}\)) \(\approx\) 53\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{SZN}\)) = \(\dfrac{SZ}{ZN}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{SZN}\)) = \(\dfrac{1{,}6}{7{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{SZN}\) =ArcCos(\(\dfrac{1{,}6}{7{,}9}\)) \(\approx\)78\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{GNT}\)) = \(\dfrac{GT}{TN}\)
\par\addvspace{.5em}
 D'où sin(40\textdegree{}) = \(\dfrac{GT}{9{,}7}\)

On a donc GT = 9,7 \(\times\) sin(40\textdegree{}) \(\approx\) 6,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{FDW}\)) = \(\dfrac{FW}{FD}\)
\par\addvspace{.5em}
d'où tan(68\textdegree{}) =\(\dfrac{FW}{9{,}9}\)
\par\addvspace{.5em}
On a donc FW = 9,9 \(\times\) tan(68\textdegree{}) \(\approx\) 24,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.303,0.792) {28°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{LGA}\)) = \(\dfrac{LG}{AG}\)
\par\addvspace{.5em}
  d'où    cos(28\textdegree{}) =  \(\dfrac{5{,}9}{AG}\)
\par\addvspace{.5em}
 On a donc AG =  \(\dfrac{5{,}9}{\cos(28^{\circ})}\) \(\approx\) 6,7 cm\end{document}
