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\begin{document}

\sloppy

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

 Dans le triangle MLZ rectangle en M, on sait que :
\begin{itemize}
\item ML=2,8 cm
\item \(\widehat{MLZ}\) = 64\textdegree{}
\end{itemize}

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

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

 Dans le triangle NBM rectangle en N, on sait que :
\begin{itemize}
\item BM=3,2 cm
\item \(\widehat{BMN}\) =  22\textdegree{}
\end{itemize}

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

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

 Dans le triangle AMR rectangle en A, on sait que :
\begin{itemize}
\item AR=5 cm
\item MR=8,8 cm
\end{itemize}

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

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

 Dans le triangle LMD rectangle en L, on sait que :
\begin{itemize}
\item LM=2,2 cm
\item \(\widehat{MDL}\) =  10\textdegree{}
\end{itemize}

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

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

 Dans le triangle JPB rectangle en J, on sait que :
\begin{itemize}
\item JB=5,5 cm
\item PB=6,9 cm
\end{itemize}

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

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

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

cos(\(\widehat{MLZ}\)) = \(\dfrac{ML}{LZ}\)
\par\addvspace{.5em}
 d'où cos(64\textdegree{}) = \(\dfrac{2{,}8}{LZ}\)
\par\addvspace{.5em}
On a donc LZ = \(\dfrac{2{,}8}{\cos(64^{\circ})}\)\(\approx\) 6,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{NMB}\)) = \(\dfrac{NB}{BM}\)
\par\addvspace{.5em}
 D'où sin(22\textdegree{}) = \(\dfrac{NB}{3{,}2}\)

On a donc NB = 3,2 \(\times\) sin(22\textdegree{}) \(\approx\) 1,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{ARM}\)) = \(\dfrac{AR}{MR}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{ARM}\)) =  \(\dfrac{5}{8{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{ARM}\) = Arccos (\(\dfrac{5}{8{,}8}\)) \(\approx\) 55\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{LDM}\)) = \(\dfrac{LM}{LD}\)
\par\addvspace{.5em}
d'où  tan(10\textdegree{}) =  \(\dfrac{2{,}2}{LD}\)
\par\addvspace{.5em}
On a donc LD = \(\dfrac{2{,}2}{\tan(10^{\circ})}\) \(\approx\) 12,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{JPB}\)) =  \(\dfrac{JB}{PB}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{JPB}\)) = \(\dfrac{5{,}5}{6{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{JPB}\) =  ArcSin(\(\dfrac{5{,}5}{6{,}9}\)) \(\approx\) 53\textdegree{}\end{document}
