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\begin{document}

\sloppy

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

 Dans le triangle VNC rectangle en V, on sait que :
\begin{itemize}
\item NC=4,9 cm
\item \(\widehat{NCV}\) =  25\textdegree{}
\end{itemize}

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

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

 Dans le triangle VAL rectangle en V, on sait que :
\begin{itemize}
\item VA=1,6 cm
\item AL=8 cm
\end{itemize}

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

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

 Dans le triangle ARW rectangle en A, on sait que :
\begin{itemize}
\item AW=8,3 cm
\item \(\widehat{RWA}\) =  43\textdegree{}
\end{itemize}

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

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

 Dans le triangle WHJ rectangle en W, on sait que :
\begin{itemize}
\item WH=3,2 cm
\item WJ=5,7 cm
\end{itemize}

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

 Dans le triangle HLN rectangle en H, on sait que :
\begin{itemize}
\item HL=7,7 cm
\item \(\widehat{HLN}\) = 50\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{VCN}\)) = \(\dfrac{VC}{NC}\)
\par\addvspace{.5em}
  d'où    cos(25\textdegree{}) =  \(\dfrac{VC}{4{,}9}\)

On a donc VC = 4,9 \(\times\) cos(25\textdegree{}) \(\approx\) 4{,}4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{VLA}\)) = \(\dfrac{VA}{AL}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{VLA}\)) =\(\dfrac{1{,}6}{8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{VLA}\) = ArcSin(\(\dfrac{1{,}6}{8}\)) \(\approx\) 12\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{AWR}\)) = \(\dfrac{AR}{AW}\)
\par\addvspace{.5em}
d'où  tan(43\textdegree{}) =  \(\dfrac{AR}{8{,}3}\)
\par\addvspace{.5em}
On a donc AR =  8,3 \(\times\) tan(43\textdegree{})  \(\approx\) 7,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{WHJ}\)) = \(\dfrac{WJ}{WH}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{WHJ}\)) = \(\dfrac{5{,}7}{3{,}2}\)
\par\addvspace{.5em}
 On a donc WHJ = ArcTan(\(\dfrac{5{,}7}{3{,}2}\)) \(\approx\) 61\textdegree{}
\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 (2.113,2.729) {50°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{HLN}\)) = \(\dfrac{HL}{LN}\)
\par\addvspace{.5em}
 d'où cos(50\textdegree{}) = \(\dfrac{7{,}7}{LN}\)
\par\addvspace{.5em}
On a donc LN = \(\dfrac{7{,}7}{\cos(50^{\circ})}\)\(\approx\) 12 cm\end{document}
