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\begin{document}

\sloppy

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

 Dans le triangle KLJ rectangle en K, on sait que :
\begin{itemize}
\item KL=6,3 cm
\item \(\widehat{KLJ}\) = 76\textdegree{}
\end{itemize}

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

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

 Dans le triangle RSF rectangle en R, on sait que :
\begin{itemize}
\item SF=5,5 cm
\item \(\widehat{SFR}\) =  18\textdegree{}
\end{itemize}

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

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

 Dans le triangle BLH rectangle en B, on sait que :
\begin{itemize}
\item BL=2,3 cm
\item BH=5,7 cm
\end{itemize}

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

 Dans le triangle LWT rectangle en L, on sait que :
\begin{itemize}
\item LW=2,4 cm
\item LT=3,6 cm
\end{itemize}

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

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

 Dans le triangle RLZ rectangle en R, on sait que :
\begin{itemize}
\item RL=6,2 cm
\item \(\widehat{RLZ}\) = 77\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{KLJ}\)) = \(\dfrac{KL}{LJ}\)
\par\addvspace{.5em}
 d'où cos(76\textdegree{}) = \(\dfrac{6{,}3}{LJ}\)
\par\addvspace{.5em}
On a donc LJ = \(\dfrac{6{,}3}{\cos(76^{\circ})}\)\(\approx\) 26 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{RFS}\)) = \(\dfrac{RS}{SF}\)
\par\addvspace{.5em}
 D'où sin(18\textdegree{}) = \(\dfrac{RS}{5{,}5}\)

On a donc RS = 5,5 \(\times\) sin(18\textdegree{}) \(\approx\) 1,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{BLH}\)) = \(\dfrac{BH}{BL}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{BLH}\)) = \(\dfrac{5{,}7}{2{,}3}\)
\par\addvspace{.5em}
 On a donc BLH = ArcTan(\(\dfrac{5{,}7}{2{,}3}\)) \(\approx\) 68\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{LTW}\)) = \(\dfrac{LW}{LT}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{LTW}\)) =  \(\dfrac{2{,}4}{3{,}6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{LTW}\) =  ArcTan(\(\dfrac{2{,}4}{3{,}6}\)) \(\approx\) 34\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.101) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.465,1.977) {77°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{RLZ}\)) = \(\dfrac{RZ}{RL}\)
\par\addvspace{.5em}
d'où tan(77\textdegree{}) =\(\dfrac{RZ}{6{,}2}\)
\par\addvspace{.5em}
On a donc RZ = 6,2 \(\times\) tan(77\textdegree{}) \(\approx\) 26,9 cm\end{document}
