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\begin{document}

\sloppy

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

 Dans le triangle CWS rectangle en C, on sait que :
\begin{itemize}
\item WS=8,4 cm
\item \(\widehat{WSC}\) =  35\textdegree{}
\end{itemize}

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

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

 Dans le triangle MFP rectangle en M, on sait que :
\begin{itemize}
\item MF=9,5 cm
\item \(\widehat{MFP}\) = 75\textdegree{}
\end{itemize}

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

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

 Dans le triangle MZH rectangle en M, on sait que :
\begin{itemize}
\item MH=5,4 cm
\item ZH=9,8 cm
\end{itemize}

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

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

 Dans le triangle LAM rectangle en L, on sait que :
\begin{itemize}
\item LM=4,9 cm
\item AM=8,7 cm
\end{itemize}

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

 Dans le triangle FVB rectangle en F, on sait que :
\begin{itemize}
\item FB=2,6 cm
\item \(\widehat{VBF}\) =  22\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{CSW}\)) = \(\dfrac{CS}{WS}\)
\par\addvspace{.5em}
  d'où    cos(35\textdegree{}) =  \(\dfrac{CS}{8{,}4}\)

On a donc CS = 8,4 \(\times\) cos(35\textdegree{}) \(\approx\) 6{,}9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{MFP}\)) = \(\dfrac{MF}{FP}\)
\par\addvspace{.5em}
 d'où cos(75\textdegree{}) = \(\dfrac{9{,}5}{FP}\)
\par\addvspace{.5em}
On a donc FP = \(\dfrac{9{,}5}{\cos(75^{\circ})}\)\(\approx\) 36,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{MHZ}\)) = \(\dfrac{MH}{ZH}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{MHZ}\)) =  \(\dfrac{5{,}4}{9{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{MHZ}\) = Arccos (\(\dfrac{5{,}4}{9{,}8}\)) \(\approx\) 57\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{LAM}\)) =  \(\dfrac{LM}{AM}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{LAM}\)) = \(\dfrac{4{,}9}{8{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{LAM}\) =  ArcSin(\(\dfrac{4{,}9}{8{,}7}\)) \(\approx\) 34\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{FBV}\)) = \(\dfrac{FV}{FB}\)
\par\addvspace{.5em}
d'où  tan(22\textdegree{}) =  \(\dfrac{FV}{2{,}6}\)
\par\addvspace{.5em}
On a donc FV =  2,6 \(\times\) tan(22\textdegree{})  \(\approx\) 1,1 cm\end{document}
