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\begin{document}

\sloppy

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

 Dans le triangle PAF rectangle en P, on sait que :
\begin{itemize}
\item AF=9,4 cm
\item \(\widehat{PAF}\) = 48\textdegree{}
\end{itemize}

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

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

 Dans le triangle GHB rectangle en G, on sait que :
\begin{itemize}
\item GB=3,6 cm
\item \(\widehat{HBG}\) =  13\textdegree{}
\end{itemize}

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

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

 Dans le triangle RPB rectangle en R, on sait que :
\begin{itemize}
\item RP=1,7 cm
\item PB=8,7 cm
\end{itemize}

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

 Dans le triangle LBM rectangle en L, on sait que :
\begin{itemize}
\item LB=1,1 cm
\item \(\widehat{LBM}\) = 56\textdegree{}
\end{itemize}

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

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

 Dans le triangle AJP rectangle en A, on sait que :
\begin{itemize}
\item AP=5 cm
\item JP=7 cm
\end{itemize}

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

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

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

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

cos(\(\widehat{PAF}\)) = \(\dfrac{PA}{AF}\)
\par\addvspace{.5em}
d'où cos(48\textdegree{}) = \(\dfrac{PA}{9{,}4}\)

On a donc PA = 9,4 \(\times\) cos(48\textdegree{}) \(\approx\) 6,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{GBH}\)) = \(\dfrac{GB}{HB}\)
\par\addvspace{.5em}
  d'où    cos(13\textdegree{}) =  \(\dfrac{3{,}6}{HB}\)
\par\addvspace{.5em}
 On a donc HB =  \(\dfrac{3{,}6}{\cos(13^{\circ})}\) \(\approx\) 3,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{RPB}\)) = \(\dfrac{RP}{PB}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{RPB}\)) = \(\dfrac{1{,}7}{8{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{RPB}\) =ArcCos(\(\dfrac{1{,}7}{8{,}7}\)) \(\approx\)79\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{LBM}\)) = \(\dfrac{LM}{LB}\)
\par\addvspace{.5em}
d'où tan(56\textdegree{}) =\(\dfrac{LM}{1{,}1}\)
\par\addvspace{.5em}
On a donc LM = 1,1 \(\times\) tan(56\textdegree{}) \(\approx\) 1,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

cos(\(\widehat{APJ}\)) = \(\dfrac{AP}{JP}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{APJ}\)) =  \(\dfrac{5}{7}\)
\par\addvspace{.5em}
On a donc \(\widehat{APJ}\) = Arccos (\(\dfrac{5}{7}\)) \(\approx\) 44\textdegree{} \end{document}
