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\begin{document}

\sloppy

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

 Dans le triangle APR rectangle en A, on sait que :
\begin{itemize}
\item AR=6,2 cm
\item PR=9,8 cm
\end{itemize}

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

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

 Dans le triangle FLB rectangle en F, on sait que :
\begin{itemize}
\item FL=9,8 cm
\item \(\widehat{FLB}\) = 58\textdegree{}
\end{itemize}

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

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

 Dans le triangle HLS rectangle en H, on sait que :
\begin{itemize}
\item HL=2,5 cm
\item HS=6,6 cm
\end{itemize}

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

 Dans le triangle LKM rectangle en L, on sait que :
\begin{itemize}
\item LM=9,3 cm
\item \(\widehat{KML}\) =  29\textdegree{}
\end{itemize}

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

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

 Dans le triangle RPJ rectangle en R, on sait que :
\begin{itemize}
\item RP=9,2 cm
\item \(\widehat{RPJ}\) = 48\textdegree{}
\end{itemize}

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

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

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

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {6,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.041,0.819) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{ARP}\)) = \(\dfrac{AR}{PR}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{ARP}\)) =  \(\dfrac{6{,}2}{9{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{ARP}\) = Arccos (\(\dfrac{6{,}2}{9{,}8}\)) \(\approx\) 51\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{FLB}\)) = \(\dfrac{FL}{LB}\)
\par\addvspace{.5em}
 d'où cos(58\textdegree{}) = \(\dfrac{9{,}8}{LB}\)
\par\addvspace{.5em}
On a donc LB = \(\dfrac{9{,}8}{\cos(58^{\circ})}\)\(\approx\) 18,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{LMK}\)) = \(\dfrac{LK}{LM}\)
\par\addvspace{.5em}
d'où  tan(29\textdegree{}) =  \(\dfrac{LK}{9{,}3}\)
\par\addvspace{.5em}
On a donc LK =  9,3 \(\times\) tan(29\textdegree{})  \(\approx\) 5,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.216,2.724) {48°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{RPJ}\)) = \(\dfrac{RJ}{RP}\)
\par\addvspace{.5em}
d'où tan(48\textdegree{}) =\(\dfrac{RJ}{9{,}2}\)
\par\addvspace{.5em}
On a donc RJ = 9,2 \(\times\) tan(48\textdegree{}) \(\approx\) 10,2 cm\end{document}
