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\begin{document}

\sloppy

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

 Dans le triangle NDB rectangle en N, on sait que :
\begin{itemize}
\item ND=2,1 cm
\item NB=3,6 cm
\end{itemize}

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

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

 Dans le triangle RVM rectangle en R, on sait que :
\begin{itemize}
\item RV=2,7 cm
\item VM=7 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{RVM}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle HNW rectangle en H, on sait que :
\begin{itemize}
\item HN=2,3 cm
\item \(\widehat{HNW}\) = 51\textdegree{}
\end{itemize}

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

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

 Dans le triangle GCH rectangle en G, on sait que :
\begin{itemize}
\item GH=8,1 cm
\item \(\widehat{GCH}\) = 56\textdegree{}
\end{itemize}

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

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

 Dans le triangle BFT rectangle en B, on sait que :
\begin{itemize}
\item BT=8,7 cm
\item \(\widehat{FTB}\) =  27\textdegree{}
\end{itemize}

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

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

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

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.256,0.707) {?};
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 Dans le triangle NDB rectangle en N, on cherche une relation entre l'angle aigu   \(\widehat{NBD}\) son coté opposé et son coté adjacent.

\par\addvspace{.5em}
tan(\(\widehat{NBD}\)) = \(\dfrac{ND}{NB}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{NBD}\)) =  \(\dfrac{2{,}1}{3{,}6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{NBD}\) =  ArcTan(\(\dfrac{2{,}1}{3{,}6}\)) \(\approx\) 30\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{RVM}\)) = \(\dfrac{RV}{VM}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{RVM}\)) = \(\dfrac{2{,}7}{7}\)
\par\addvspace{.5em}
On a donc \(\widehat{RVM}\) =ArcCos(\(\dfrac{2{,}7}{7}\)) \(\approx\)67\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{HNW}\)) = \(\dfrac{HW}{HN}\)
\par\addvspace{.5em}
d'où tan(51\textdegree{}) =\(\dfrac{HW}{2{,}3}\)
\par\addvspace{.5em}
On a donc HW = 2,3 \(\times\) tan(51\textdegree{}) \(\approx\) 2,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.737,2.745) {56°};
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 Dans le triangle GCH rectangle en G, on cherche une relation entre l'angle aigu   \(\widehat{GCH}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{GCH}\)) =  \(\dfrac{GH}{CH}\)
\par\addvspace{.5em}
 d'où sin(56\textdegree{}) = \(\dfrac{8{,}1}{CH}\)
\par\addvspace{.5em}
 On a donc CH = \(\dfrac{8{,}1}{\sin(56^{\circ})}\) \(\approx\) 9,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {?};
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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.302,0.841) {27°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{BTF}\)) = \(\dfrac{BF}{BT}\)
\par\addvspace{.5em}
d'où  tan(27\textdegree{}) =  \(\dfrac{BF}{8{,}7}\)
\par\addvspace{.5em}
On a donc BF =  8,7 \(\times\) tan(27\textdegree{})  \(\approx\) 4,4 cm\end{document}
