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\begin{document}

\sloppy

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

 Dans le triangle BFC rectangle en B, on sait que :
\begin{itemize}
\item BF=1 cm
\item FC=10 cm
\end{itemize}

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

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

 Dans le triangle WAP rectangle en W, on sait que :
\begin{itemize}
\item WA=8,2 cm
\item \(\widehat{APW}\) =  24\textdegree{}
\end{itemize}

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

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

 Dans le triangle RNV rectangle en R, on sait que :
\begin{itemize}
\item RV=1,4 cm
\item \(\widehat{RNV}\) = 77\textdegree{}
\end{itemize}

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

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

 Dans le triangle MLV rectangle en M, on sait que :
\begin{itemize}
\item LV=8,3 cm
\item \(\widehat{LVM}\) =  21\textdegree{}
\end{itemize}

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

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

 Dans le triangle TRL rectangle en T, on sait que :
\begin{itemize}
\item TR=1 cm
\item RL=9,4 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{TRL}\). \newpage
\section*{\textcolor{red}{Correction}}

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

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

\par\addvspace{.5em}
sin(\(\widehat{BCF}\)) = \(\dfrac{BF}{FC}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{BCF}\)) =\(\dfrac{1}{10}\)
\par\addvspace{.5em}
 On a donc \(\widehat{BCF}\) = ArcSin(\(\dfrac{1}{10}\)) \(\approx\) 6\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{WPA}\)) = \(\dfrac{WA}{WP}\)
\par\addvspace{.5em}
d'où  tan(24\textdegree{}) =  \(\dfrac{8{,}2}{WP}\)
\par\addvspace{.5em}
On a donc WP = \(\dfrac{8{,}2}{\tan(24^{\circ})}\) \(\approx\) 18,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{RNV}\)) =  \(\dfrac{RV}{NV}\)
\par\addvspace{.5em}
 d'où sin(77\textdegree{}) = \(\dfrac{1{,}4}{NV}\)
\par\addvspace{.5em}
 On a donc NV = \(\dfrac{1{,}4}{\sin(77^{\circ})}\) \(\approx\) 1,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{MVL}\)) = \(\dfrac{ML}{LV}\)
\par\addvspace{.5em}
 D'où sin(21\textdegree{}) = \(\dfrac{ML}{8{,}3}\)

On a donc ML = 8,3 \(\times\) sin(21\textdegree{}) \(\approx\) 3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.496,1.700) {?};
\end{tikzpicture}\end{center}


 Dans le triangle TRL rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TRL}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{TRL}\)) = \(\dfrac{TR}{RL}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{TRL}\)) = \(\dfrac{1}{9{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{TRL}\) =ArcCos(\(\dfrac{1}{9{,}4}\)) \(\approx\)84\textdegree{}\end{document}
