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\begin{document}

\sloppy

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

 Dans le triangle NVT rectangle en N, on sait que :
\begin{itemize}
\item VT=2,7 cm
\item \(\widehat{NVT}\) = 63\textdegree{}
\end{itemize}

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

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

 Dans le triangle GVC rectangle en G, on sait que :
\begin{itemize}
\item GV=3,2 cm
\item VC=8,3 cm
\end{itemize}

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

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

 Dans le triangle CST rectangle en C, on sait que :
\begin{itemize}
\item CS=5,5 cm
\item \(\widehat{CST}\) = 53\textdegree{}
\end{itemize}

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

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

 Dans le triangle JWA rectangle en J, on sait que :
\begin{itemize}
\item JW=1,8 cm
\item WA=9,9 cm
\end{itemize}

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

 Dans le triangle MRS rectangle en M, on sait que :
\begin{itemize}
\item MR=1 cm
\item \(\widehat{RSM}\) =  44\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{NVT}\)) = \(\dfrac{NV}{VT}\)
\par\addvspace{.5em}
d'où cos(63\textdegree{}) = \(\dfrac{NV}{2{,}7}\)

On a donc NV = 2,7 \(\times\) cos(63\textdegree{}) \(\approx\) 1,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{GCV}\)) = \(\dfrac{GV}{VC}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{GCV}\)) =\(\dfrac{3{,}2}{8{,}3}\)
\par\addvspace{.5em}
 On a donc \(\widehat{GCV}\) = ArcSin(\(\dfrac{3{,}2}{8{,}3}\)) \(\approx\) 23\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{CST}\)) = \(\dfrac{CS}{ST}\)
\par\addvspace{.5em}
 d'où cos(53\textdegree{}) = \(\dfrac{5{,}5}{ST}\)
\par\addvspace{.5em}
On a donc ST = \(\dfrac{5{,}5}{\cos(53^{\circ})}\)\(\approx\) 9,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{JWA}\)) = \(\dfrac{JW}{WA}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{JWA}\)) = \(\dfrac{1{,}8}{9{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{JWA}\) =ArcCos(\(\dfrac{1{,}8}{9{,}9}\)) \(\approx\)80\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.332,0.783) {44°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{MSR}\)) = \(\dfrac{MR}{MS}\)
\par\addvspace{.5em}
d'où  tan(44\textdegree{}) =  \(\dfrac{1}{MS}\)
\par\addvspace{.5em}
On a donc MS = \(\dfrac{1}{\tan(44^{\circ})}\) \(\approx\) 1 cm\end{document}
