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\begin{document}

\sloppy

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

 Dans le triangle JDV rectangle en J, on sait que :
\begin{itemize}
\item JD=1,9 cm
\item DV=6,7 cm
\end{itemize}

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

 Dans le triangle MFW rectangle en M, on sait que :
\begin{itemize}
\item MW=2,5 cm
\item \(\widehat{MFW}\) = 55\textdegree{}
\end{itemize}

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

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

 Dans le triangle LCK rectangle en L, on sait que :
\begin{itemize}
\item LC=2,6 cm
\item \(\widehat{LCK}\) = 75\textdegree{}
\end{itemize}

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

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

 Dans le triangle DPM rectangle en D, on sait que :
\begin{itemize}
\item PM=7,7 cm
\item \(\widehat{DPM}\) = 47\textdegree{}
\end{itemize}

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

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

 Dans le triangle CNT rectangle en C, on sait que :
\begin{itemize}
\item CN=1,4 cm
\item NT=9,3 cm
\end{itemize}

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

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

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

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

cos(\(\widehat{JDV}\)) = \(\dfrac{JD}{DV}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{JDV}\)) = \(\dfrac{1{,}9}{6{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{JDV}\) =ArcCos(\(\dfrac{1{,}9}{6{,}7}\)) \(\approx\)74\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{MFW}\)) = \(\dfrac{MW}{MF}\)
\par\addvspace{.5em}
tan(55\textdegree{}) = \(\dfrac{2{,}5}{MF}\)
\par\addvspace{.5em}
 On a donc MF = \(\dfrac{2{,}5}{\tan(55^{\circ})}\) \(\approx\) 1,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{LCK}\)) = \(\dfrac{LC}{CK}\)
\par\addvspace{.5em}
 d'où cos(75\textdegree{}) = \(\dfrac{2{,}6}{CK}\)
\par\addvspace{.5em}
On a donc CK = \(\dfrac{2{,}6}{\cos(75^{\circ})}\)\(\approx\) 10 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{DPM}\)) =  \(\dfrac{DM}{PM}\)
\par\addvspace{.5em}
 d'où sin(47\textdegree{}) = \(\dfrac{DM}{7{,}7}\)

On a donc DM = 7,7 \(\times\) sin(47\textdegree{}) \(\approx\) 5,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.285,1.611) {?};
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 Dans le triangle CNT rectangle en C, on cherche une relation entre l'angle aigu   \(\widehat{CTN}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{CTN}\)) = \(\dfrac{CN}{NT}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{CTN}\)) =\(\dfrac{1{,}4}{9{,}3}\)
\par\addvspace{.5em}
 On a donc \(\widehat{CTN}\) = ArcSin(\(\dfrac{1{,}4}{9{,}3}\)) \(\approx\) 9\textdegree{}\end{document}
