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\begin{document}

\sloppy

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

 Dans le triangle TNK rectangle en T, on sait que :
\begin{itemize}
\item NK=6,9 cm
\item \(\widehat{NKT}\) =  11\textdegree{}
\end{itemize}

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

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

 Dans le triangle NAK rectangle en N, on sait que :
\begin{itemize}
\item NK=2,8 cm
\item \(\widehat{NAK}\) = 70\textdegree{}
\end{itemize}

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

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

 Dans le triangle LJN rectangle en L, on sait que :
\begin{itemize}
\item LN=3,5 cm
\item JN=7,4 cm
\end{itemize}

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

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

 Dans le triangle KGC rectangle en K, on sait que :
\begin{itemize}
\item KG=2,3 cm
\item KC=6 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{KGC}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle TJW rectangle en T, on sait que :
\begin{itemize}
\item JW=7,6 cm
\item \(\widehat{JWT}\) =  29\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{TKN}\)) = \(\dfrac{TN}{NK}\)
\par\addvspace{.5em}
 D'où sin(11\textdegree{}) = \(\dfrac{TN}{6{,}9}\)

On a donc TN = 6,9 \(\times\) sin(11\textdegree{}) \(\approx\) 1,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{NAK}\)) =  \(\dfrac{NK}{AK}\)
\par\addvspace{.5em}
 d'où sin(70\textdegree{}) = \(\dfrac{2{,}8}{AK}\)
\par\addvspace{.5em}
 On a donc AK = \(\dfrac{2{,}8}{\sin(70^{\circ})}\) \(\approx\) 3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{LNJ}\)) = \(\dfrac{LN}{JN}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{LNJ}\)) =  \(\dfrac{3{,}5}{7{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{LNJ}\) = Arccos (\(\dfrac{3{,}5}{7{,}4}\)) \(\approx\) 62\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{KGC}\)) = \(\dfrac{KC}{KG}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{KGC}\)) = \(\dfrac{6}{2{,}3}\)
\par\addvspace{.5em}
 On a donc KGC = ArcTan(\(\dfrac{6}{2{,}3}\)) \(\approx\) 69\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.305,0.742) {29°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{TWJ}\)) = \(\dfrac{TW}{JW}\)
\par\addvspace{.5em}
  d'où    cos(29\textdegree{}) =  \(\dfrac{TW}{7{,}6}\)

On a donc TW = 7,6 \(\times\) cos(29\textdegree{}) \(\approx\) 6{,}6 cm\end{document}
