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\begin{document}

\sloppy

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

 Dans le triangle KGA rectangle en K, on sait que :
\begin{itemize}
\item GA=9,1 cm
\item \(\widehat{KGA}\) = 71\textdegree{}
\end{itemize}

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

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

 Dans le triangle GKD rectangle en G, on sait que :
\begin{itemize}
\item GK=1,3 cm
\item KD=8 cm
\end{itemize}

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

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

 Dans le triangle HFB rectangle en H, on sait que :
\begin{itemize}
\item HF=1,9 cm
\item FB=7,5 cm
\end{itemize}

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

 Dans le triangle FLJ rectangle en F, on sait que :
\begin{itemize}
\item FJ=1,6 cm
\item \(\widehat{LJF}\) =  11\textdegree{}
\end{itemize}

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

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

 Dans le triangle TKL rectangle en T, on sait que :
\begin{itemize}
\item TL=4,8 cm
\item \(\widehat{KLT}\) =  30\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{KGA}\)) =  \(\dfrac{KA}{GA}\)
\par\addvspace{.5em}
 d'où sin(71\textdegree{}) = \(\dfrac{KA}{9{,}1}\)

On a donc KA = 9,1 \(\times\) sin(71\textdegree{}) \(\approx\) 8,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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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 l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{GDK}\)) = \(\dfrac{GK}{KD}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{GDK}\)) =\(\dfrac{1{,}3}{8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{GDK}\) = ArcSin(\(\dfrac{1{,}3}{8}\)) \(\approx\) 9\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{HFB}\)) = \(\dfrac{HF}{FB}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{HFB}\)) = \(\dfrac{1{,}9}{7{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{HFB}\) =ArcCos(\(\dfrac{1{,}9}{7{,}5}\)) \(\approx\)75\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{FJL}\)) = \(\dfrac{FJ}{LJ}\)
\par\addvspace{.5em}
  d'où    cos(11\textdegree{}) =  \(\dfrac{1{,}6}{LJ}\)
\par\addvspace{.5em}
 On a donc LJ =  \(\dfrac{1{,}6}{\cos(11^{\circ})}\) \(\approx\) 1,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.025,0.467) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.575,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {4,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.281,0.706) {30°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{TLK}\)) = \(\dfrac{TK}{TL}\)
\par\addvspace{.5em}
d'où  tan(30\textdegree{}) =  \(\dfrac{TK}{4{,}8}\)
\par\addvspace{.5em}
On a donc TK =  4,8 \(\times\) tan(30\textdegree{})  \(\approx\) 2,8 cm\end{document}
