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\begin{document}

\sloppy

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

 Dans le triangle KPT rectangle en K, on sait que :
\begin{itemize}
\item KT=5,9 cm
\item PT=7,6 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{KPT}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle HDM rectangle en H, on sait que :
\begin{itemize}
\item HD=0,4 cm
\item \(\widehat{DMH}\) =  38\textdegree{}
\end{itemize}

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

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

 Dans le triangle JLW rectangle en J, on sait que :
\begin{itemize}
\item JL=2,7 cm
\item JW=4,1 cm
\end{itemize}

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

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

 Dans le triangle DWT rectangle en D, on sait que :
\begin{itemize}
\item WT=7,6 cm
\item \(\widehat{DWT}\) = 46\textdegree{}
\end{itemize}

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

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

 Dans le triangle FWK rectangle en F, on sait que :
\begin{itemize}
\item FW=7,1 cm
\item \(\widehat{WKF}\) =  40\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [KW]. (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 (2.062,2.731) {?};
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 Dans le triangle KPT rectangle en K, on cherche une relation entre l'angle aigu   \(\widehat{KPT}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{KPT}\)) =  \(\dfrac{KT}{PT}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{KPT}\)) = \(\dfrac{5{,}9}{7{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{KPT}\) =  ArcSin(\(\dfrac{5{,}9}{7{,}6}\)) \(\approx\) 51\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{HMD}\)) = \(\dfrac{HD}{HM}\)
\par\addvspace{.5em}
d'où  tan(38\textdegree{}) =  \(\dfrac{0{,}4}{HM}\)
\par\addvspace{.5em}
On a donc HM = \(\dfrac{0{,}4}{\tan(38^{\circ})}\) \(\approx\) 0,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{JWL}\)) = \(\dfrac{JL}{JW}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{JWL}\)) =  \(\dfrac{2{,}7}{4{,}1}\)
\par\addvspace{.5em}
 On a donc \(\widehat{JWL}\) =  ArcTan(\(\dfrac{2{,}7}{4{,}1}\)) \(\approx\) 33\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.311,2.720) {46°};
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 Dans le triangle DWT rectangle en D, on cherche une relation entre l'angle aigu   \(\widehat{DWT}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{DWT}\)) = \(\dfrac{DW}{WT}\)
\par\addvspace{.5em}
d'où cos(46\textdegree{}) = \(\dfrac{DW}{7{,}6}\)

On a donc DW = 7,6 \(\times\) cos(46\textdegree{}) \(\approx\) 5,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{FKW}\)) = \(\dfrac{FW}{WK}\)
\par\addvspace{.5em}
 D'où sin(40\textdegree{}) = \(\dfrac{7{,}1}{WK}\)
\par\addvspace{.5em}
On a donc WK =\(\dfrac{7{,}1}{\sin(40^{\circ})}\) \(\approx\) 11 cm\end{document}
