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\begin{document}

\sloppy

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

 Dans le triangle MAV rectangle en M, on sait que :
\begin{itemize}
\item AV=4,2 cm
\item \(\widehat{AVM}\) =  23\textdegree{}
\end{itemize}

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

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

 Dans le triangle VAB rectangle en V, on sait que :
\begin{itemize}
\item VA=1,4 cm
\item VB=5,6 cm
\end{itemize}

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

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

 Dans le triangle VMC rectangle en V, on sait que :
\begin{itemize}
\item VC=5,4 cm
\item MC=9,6 cm
\end{itemize}

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

 Dans le triangle RTD rectangle en R, on sait que :
\begin{itemize}
\item RT=6,3 cm
\item \(\widehat{RTD}\) = 62\textdegree{}
\end{itemize}

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

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

 Dans le triangle PKV rectangle en P, on sait que :
\begin{itemize}
\item PK=0,7 cm
\item \(\widehat{PKV}\) = 48\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{MVA}\)) = \(\dfrac{MA}{AV}\)
\par\addvspace{.5em}
 D'où sin(23\textdegree{}) = \(\dfrac{MA}{4{,}2}\)

On a donc MA = 4,2 \(\times\) sin(23\textdegree{}) \(\approx\) 1,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{VBA}\)) = \(\dfrac{VA}{VB}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{VBA}\)) =  \(\dfrac{1{,}4}{5{,}6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{VBA}\) =  ArcTan(\(\dfrac{1{,}4}{5{,}6}\)) \(\approx\) 14\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{VMC}\)) =  \(\dfrac{VC}{MC}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{VMC}\)) = \(\dfrac{5{,}4}{9{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{VMC}\) =  ArcSin(\(\dfrac{5{,}4}{9{,}6}\)) \(\approx\) 34\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{RTD}\)) = \(\dfrac{RD}{RT}\)
\par\addvspace{.5em}
d'où tan(62\textdegree{}) =\(\dfrac{RD}{6{,}3}\)
\par\addvspace{.5em}
On a donc RD = 6,3 \(\times\) tan(62\textdegree{}) \(\approx\) 11,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.216,2.724) {48°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{PKV}\)) = \(\dfrac{PK}{KV}\)
\par\addvspace{.5em}
 d'où cos(48\textdegree{}) = \(\dfrac{0{,}7}{KV}\)
\par\addvspace{.5em}
On a donc KV = \(\dfrac{0{,}7}{\cos(48^{\circ})}\)\(\approx\) 1 cm\end{document}
