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\begin{document}

\sloppy

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

 Dans le triangle NCG rectangle en N, on sait que :
\begin{itemize}
\item NG=8,3 cm
\item \(\widehat{NCG}\) = 53\textdegree{}
\end{itemize}

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

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

 Dans le triangle TKA rectangle en T, on sait que :
\begin{itemize}
\item TK=8,2 cm
\item \(\widehat{TKA}\) = 61\textdegree{}
\end{itemize}

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

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

 Dans le triangle CAV rectangle en C, on sait que :
\begin{itemize}
\item AV=5,8 cm
\item \(\widehat{CAV}\) = 64\textdegree{}
\end{itemize}

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

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

 Dans le triangle LGA rectangle en L, on sait que :
\begin{itemize}
\item LA=6,2 cm
\item GA=7 cm
\end{itemize}

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

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

 Dans le triangle FSZ rectangle en F, on sait que :
\begin{itemize}
\item FS=3,1 cm
\item SZ=8,2 cm
\end{itemize}

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{NCG}\)) = \(\dfrac{NG}{NC}\)
\par\addvspace{.5em}
tan(53\textdegree{}) = \(\dfrac{8{,}3}{NC}\)
\par\addvspace{.5em}
 On a donc NC = \(\dfrac{8{,}3}{\tan(53^{\circ})}\) \(\approx\) 6,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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 Dans le triangle TKA rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TKA}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{TKA}\)) = \(\dfrac{TK}{KA}\)
\par\addvspace{.5em}
 d'où cos(61\textdegree{}) = \(\dfrac{8{,}2}{KA}\)
\par\addvspace{.5em}
On a donc KA = \(\dfrac{8{,}2}{\cos(61^{\circ})}\)\(\approx\) 16,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{CAV}\)) =  \(\dfrac{CV}{AV}\)
\par\addvspace{.5em}
 d'où sin(64\textdegree{}) = \(\dfrac{CV}{5{,}8}\)

On a donc CV = 5,8 \(\times\) sin(64\textdegree{}) \(\approx\) 5,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{LAG}\)) = \(\dfrac{LA}{GA}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{LAG}\)) =  \(\dfrac{6{,}2}{7}\)
\par\addvspace{.5em}
On a donc \(\widehat{LAG}\) = Arccos (\(\dfrac{6{,}2}{7}\)) \(\approx\) 28\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) {8,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.422,2.380) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{FSZ}\)) = \(\dfrac{FS}{SZ}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{FSZ}\)) = \(\dfrac{3{,}1}{8{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{FSZ}\) =ArcCos(\(\dfrac{3{,}1}{8{,}2}\)) \(\approx\)68\textdegree{}\end{document}
