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\begin{document}

\sloppy

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

 Dans le triangle HDL rectangle en H, on sait que :
\begin{itemize}
\item HL=4,3 cm
\item DL=8,6 cm
\end{itemize}

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

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

 Dans le triangle JCR rectangle en J, on sait que :
\begin{itemize}
\item JR=3,7 cm
\item \(\widehat{CRJ}\) =  18\textdegree{}
\end{itemize}

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

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

 Dans le triangle ALB rectangle en A, on sait que :
\begin{itemize}
\item AB=1,8 cm
\item \(\widehat{ALB}\) = 52\textdegree{}
\end{itemize}

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

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

 Dans le triangle TVB rectangle en T, on sait que :
\begin{itemize}
\item TV=8,3 cm
\item \(\widehat{TVB}\) = 72\textdegree{}
\end{itemize}

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

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

 Dans le triangle FDM rectangle en F, on sait que :
\begin{itemize}
\item FM=6,1 cm
\item DM=9,3 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{FDM}\). \newpage
\section*{\textcolor{red}{Correction}}

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

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

cos(\(\widehat{HLD}\)) = \(\dfrac{HL}{DL}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{HLD}\)) =  \(\dfrac{4{,}3}{8{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{HLD}\) = Arccos (\(\dfrac{4{,}3}{8{,}6}\)) \(\approx\) 60\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{JRC}\)) = \(\dfrac{JC}{JR}\)
\par\addvspace{.5em}
d'où  tan(18\textdegree{}) =  \(\dfrac{JC}{3{,}7}\)
\par\addvspace{.5em}
On a donc JC =  3,7 \(\times\) tan(18\textdegree{})  \(\approx\) 1,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{ALB}\)) =  \(\dfrac{AB}{LB}\)
\par\addvspace{.5em}
 d'où sin(52\textdegree{}) = \(\dfrac{1{,}8}{LB}\)
\par\addvspace{.5em}
 On a donc LB = \(\dfrac{1{,}8}{\sin(52^{\circ})}\) \(\approx\) 2,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{TVB}\)) = \(\dfrac{TB}{TV}\)
\par\addvspace{.5em}
d'où tan(72\textdegree{}) =\(\dfrac{TB}{8{,}3}\)
\par\addvspace{.5em}
On a donc TB = 8,3 \(\times\) tan(72\textdegree{}) \(\approx\) 25,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{FDM}\)) =  \(\dfrac{FM}{DM}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{FDM}\)) = \(\dfrac{6{,}1}{9{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{FDM}\) =  ArcSin(\(\dfrac{6{,}1}{9{,}3}\)) \(\approx\) 41\textdegree{}\end{document}
