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\begin{document}

\sloppy

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

 Dans le triangle GKD rectangle en G, on sait que :
\begin{itemize}
\item GD=7,6 cm
\item \(\widehat{KDG}\) =  33\textdegree{}
\end{itemize}

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

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

 Dans le triangle KJT rectangle en K, on sait que :
\begin{itemize}
\item JT=6 cm
\item \(\widehat{JTK}\) =  23\textdegree{}
\end{itemize}

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

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

 Dans le triangle NLV rectangle en N, on sait que :
\begin{itemize}
\item NL=1,2 cm
\item LV=9,8 cm
\end{itemize}

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

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

 Dans le triangle SCV rectangle en S, on sait que :
\begin{itemize}
\item SC=8,7 cm
\item \(\widehat{SCV}\) = 63\textdegree{}
\end{itemize}

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

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

 Dans le triangle RZP rectangle en R, on sait que :
\begin{itemize}
\item RZ=1,1 cm
\item ZP=9,6 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{RZP}\). \newpage
\section*{\textcolor{red}{Correction}}

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

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

\par\addvspace{.5em}
tan(\(\widehat{GDK}\)) = \(\dfrac{GK}{GD}\)
\par\addvspace{.5em}
d'où  tan(33\textdegree{}) =  \(\dfrac{GK}{7{,}6}\)
\par\addvspace{.5em}
On a donc GK =  7,6 \(\times\) tan(33\textdegree{})  \(\approx\) 4,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{KTJ}\)) = \(\dfrac{KT}{JT}\)
\par\addvspace{.5em}
  d'où    cos(23\textdegree{}) =  \(\dfrac{KT}{6}\)

On a donc KT = 6 \(\times\) cos(23\textdegree{}) \(\approx\) 5{,}5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{NVL}\)) = \(\dfrac{NL}{LV}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{NVL}\)) =\(\dfrac{1{,}2}{9{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{NVL}\) = ArcSin(\(\dfrac{1{,}2}{9{,}8}\)) \(\approx\) 7\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{SCV}\)) = \(\dfrac{SC}{CV}\)
\par\addvspace{.5em}
 d'où cos(63\textdegree{}) = \(\dfrac{8{,}7}{CV}\)
\par\addvspace{.5em}
On a donc CV = \(\dfrac{8{,}7}{\cos(63^{\circ})}\)\(\approx\) 19,2 cm
\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) {1,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.494,1.718) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{RZP}\)) = \(\dfrac{RZ}{ZP}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{RZP}\)) = \(\dfrac{1{,}1}{9{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{RZP}\) =ArcCos(\(\dfrac{1{,}1}{9{,}6}\)) \(\approx\)83\textdegree{}\end{document}
