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\begin{document}

\sloppy

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

 Dans le triangle TGS rectangle en T, on sait que :
\begin{itemize}
\item TG=1,1 cm
\item GS=6,8 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{TGS}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle BKT rectangle en B, on sait que :
\begin{itemize}
\item BT=10 cm
\item \(\widehat{BKT}\) = 56\textdegree{}
\end{itemize}

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

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

 Dans le triangle BSA rectangle en B, on sait que :
\begin{itemize}
\item BS=1,8 cm
\item SA=8,8 cm
\end{itemize}

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

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

 Dans le triangle DLF rectangle en D, on sait que :
\begin{itemize}
\item DL=3,3 cm
\item \(\widehat{LFD}\) =  17\textdegree{}
\end{itemize}

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

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

 Dans le triangle TJN rectangle en T, on sait que :
\begin{itemize}
\item JN=2,9 cm
\item \(\widehat{JNT}\) =  22\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{TGS}\)) = \(\dfrac{TG}{GS}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{TGS}\)) = \(\dfrac{1{,}1}{6{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{TGS}\) =ArcCos(\(\dfrac{1{,}1}{6{,}8}\)) \(\approx\)81\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{BKT}\)) = \(\dfrac{BT}{BK}\)
\par\addvspace{.5em}
tan(56\textdegree{}) = \(\dfrac{10}{BK}\)
\par\addvspace{.5em}
 On a donc BK = \(\dfrac{10}{\tan(56^{\circ})}\) \(\approx\) 6,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{BAS}\)) = \(\dfrac{BS}{SA}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{BAS}\)) =\(\dfrac{1{,}8}{8{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{BAS}\) = ArcSin(\(\dfrac{1{,}8}{8{,}8}\)) \(\approx\) 12\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{DFL}\)) = \(\dfrac{DL}{LF}\)
\par\addvspace{.5em}
 D'où sin(17\textdegree{}) = \(\dfrac{3{,}3}{LF}\)
\par\addvspace{.5em}
On a donc LF =\(\dfrac{3{,}3}{\sin(17^{\circ})}\) \(\approx\) 11,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.677) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {2,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.296,1.071) {22°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{TNJ}\)) = \(\dfrac{TN}{JN}\)
\par\addvspace{.5em}
  d'où    cos(22\textdegree{}) =  \(\dfrac{TN}{2{,}9}\)

On a donc TN = 2,9 \(\times\) cos(22\textdegree{}) \(\approx\) 2{,}7 cm\end{document}
