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\begin{document}

\sloppy

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

 Dans le triangle WVN rectangle en W, on sait que :
\begin{itemize}
\item WN=9,9 cm
\item \(\widehat{WVN}\) = 76\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 NHJ rectangle en N, on sait que :
\begin{itemize}
\item HJ=2,8 cm
\item \(\widehat{HJN}\) =  26\textdegree{}
\end{itemize}

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

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

 Dans le triangle VZH rectangle en V, on sait que :
\begin{itemize}
\item VH=5,4 cm
\item ZH=9,7 cm
\end{itemize}

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

 Dans le triangle WTA rectangle en W, on sait que :
\begin{itemize}
\item WT=1,7 cm
\item WA=3,5 cm
\end{itemize}

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

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

 Dans le triangle CDB rectangle en C, on sait que :
\begin{itemize}
\item CB=3,9 cm
\item \(\widehat{DBC}\) =  40\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{WVN}\)) =  \(\dfrac{WN}{VN}\)
\par\addvspace{.5em}
 d'où sin(76\textdegree{}) = \(\dfrac{9{,}9}{VN}\)
\par\addvspace{.5em}
 On a donc VN = \(\dfrac{9{,}9}{\sin(76^{\circ})}\) \(\approx\) 10,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{NJH}\)) = \(\dfrac{NJ}{HJ}\)
\par\addvspace{.5em}
  d'où    cos(26\textdegree{}) =  \(\dfrac{NJ}{2{,}8}\)

On a donc NJ = 2,8 \(\times\) cos(26\textdegree{}) \(\approx\) 2{,}5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{WAT}\)) = \(\dfrac{WT}{WA}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{WAT}\)) =  \(\dfrac{1{,}7}{3{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{WAT}\) =  ArcTan(\(\dfrac{1{,}7}{3{,}5}\)) \(\approx\) 26\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {3,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.542,0.761) {40°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{CBD}\)) = \(\dfrac{CD}{CB}\)
\par\addvspace{.5em}
d'où  tan(40\textdegree{}) =  \(\dfrac{CD}{3{,}9}\)
\par\addvspace{.5em}
On a donc CD =  3,9 \(\times\) tan(40\textdegree{})  \(\approx\) 3,3 cm\end{document}
