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\begin{document}

\sloppy

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

 Dans le triangle JPM rectangle en J, on sait que :
\begin{itemize}
\item JP=8,7 cm
\item \(\widehat{JPM}\) = 70\textdegree{}
\end{itemize}

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

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

 Dans le triangle CSW rectangle en C, on sait que :
\begin{itemize}
\item CS=4,3 cm
\item \(\widehat{CSW}\) = 59\textdegree{}
\end{itemize}

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

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

 Dans le triangle KPW rectangle en K, on sait que :
\begin{itemize}
\item PW=8,3 cm
\item \(\widehat{KPW}\) = 76\textdegree{}
\end{itemize}

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

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

 Dans le triangle FWN rectangle en F, on sait que :
\begin{itemize}
\item FN=5,6 cm
\item WN=9,2 cm
\end{itemize}

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

 Dans le triangle KVH rectangle en K, on sait que :
\begin{itemize}
\item KV=3,2 cm
\item KH=4,3 cm
\end{itemize}

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

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

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

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

cos(\(\widehat{JPM}\)) = \(\dfrac{JP}{PM}\)
\par\addvspace{.5em}
 d'où cos(70\textdegree{}) = \(\dfrac{8{,}7}{PM}\)
\par\addvspace{.5em}
On a donc PM = \(\dfrac{8{,}7}{\cos(70^{\circ})}\)\(\approx\) 25,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{CSW}\)) = \(\dfrac{CW}{CS}\)
\par\addvspace{.5em}
d'où tan(59\textdegree{}) =\(\dfrac{CW}{4{,}3}\)
\par\addvspace{.5em}
On a donc CW = 4,3 \(\times\) tan(59\textdegree{}) \(\approx\) 7,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{KPW}\)) = \(\dfrac{KP}{PW}\)
\par\addvspace{.5em}
d'où cos(76\textdegree{}) = \(\dfrac{KP}{8{,}3}\)

On a donc KP = 8,3 \(\times\) cos(76\textdegree{}) \(\approx\) 2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.640,2.702) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{FWN}\)) =  \(\dfrac{FN}{WN}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{FWN}\)) = \(\dfrac{5{,}6}{9{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{FWN}\) =  ArcSin(\(\dfrac{5{,}6}{9{,}2}\)) \(\approx\) 37\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.748,0.743) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{KHV}\)) = \(\dfrac{KV}{KH}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{KHV}\)) =  \(\dfrac{3{,}2}{4{,}3}\)
\par\addvspace{.5em}
 On a donc \(\widehat{KHV}\) =  ArcTan(\(\dfrac{3{,}2}{4{,}3}\)) \(\approx\) 37\textdegree{}\end{document}
