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\begin{document}

\sloppy

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

 Dans le triangle PBJ rectangle en P, on sait que :
\begin{itemize}
\item BJ=5,5 cm
\item \(\widehat{BJP}\) =  36\textdegree{}
\end{itemize}

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

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

 Dans le triangle WLA rectangle en W, on sait que :
\begin{itemize}
\item LA=1,5 cm
\item \(\widehat{LAW}\) =  21\textdegree{}
\end{itemize}

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

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

 Dans le triangle PDJ rectangle en P, on sait que :
\begin{itemize}
\item PJ=3,5 cm
\item DJ=7,5 cm
\end{itemize}

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

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

 Dans le triangle PLM rectangle en P, on sait que :
\begin{itemize}
\item PL=1,6 cm
\item LM=9,8 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{PLM}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle MKT rectangle en M, on sait que :
\begin{itemize}
\item MK=3,6 cm
\item \(\widehat{KTM}\) =  20\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{PJB}\)) = \(\dfrac{PB}{BJ}\)
\par\addvspace{.5em}
 D'où sin(36\textdegree{}) = \(\dfrac{PB}{5{,}5}\)

On a donc PB = 5,5 \(\times\) sin(36\textdegree{}) \(\approx\) 3,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{WAL}\)) = \(\dfrac{WA}{LA}\)
\par\addvspace{.5em}
  d'où    cos(21\textdegree{}) =  \(\dfrac{WA}{1{,}5}\)

On a donc WA = 1,5 \(\times\) cos(21\textdegree{}) \(\approx\) 1{,}4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{PJD}\)) = \(\dfrac{PJ}{DJ}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{PJD}\)) =  \(\dfrac{3{,}5}{7{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{PJD}\) = Arccos (\(\dfrac{3{,}5}{7{,}5}\)) \(\approx\) 62\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{PLM}\)) = \(\dfrac{PL}{LM}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{PLM}\)) = \(\dfrac{1{,}6}{9{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{PLM}\) =ArcCos(\(\dfrac{1{,}6}{9{,}8}\)) \(\approx\)81\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{MTK}\)) = \(\dfrac{MK}{KT}\)
\par\addvspace{.5em}
 D'où sin(20\textdegree{}) = \(\dfrac{3{,}6}{KT}\)
\par\addvspace{.5em}
On a donc KT =\(\dfrac{3{,}6}{\sin(20^{\circ})}\) \(\approx\) 10,5 cm\end{document}
