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\begin{document}

\sloppy

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

 Dans le triangle TNA rectangle en T, on sait que :
\begin{itemize}
\item TA=3,8 cm
\item NA=8,7 cm
\end{itemize}

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

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

 Dans le triangle VMT rectangle en V, on sait que :
\begin{itemize}
\item VM=3,3 cm
\item VT=4,8 cm
\end{itemize}

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

 Dans le triangle NRL rectangle en N, on sait que :
\begin{itemize}
\item NR=7 cm
\item \(\widehat{RLN}\) =  20\textdegree{}
\end{itemize}

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

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

 Dans le triangle BDH rectangle en B, on sait que :
\begin{itemize}
\item BH=2,2 cm
\item \(\widehat{DHB}\) =  45\textdegree{}
\end{itemize}

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

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

 Dans le triangle MSP rectangle en M, on sait que :
\begin{itemize}
\item SP=4,7 cm
\item \(\widehat{MSP}\) = 77\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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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.615,0.887) {?};
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 Dans le triangle TNA rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TAN}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{TAN}\)) = \(\dfrac{TA}{NA}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{TAN}\)) =  \(\dfrac{3{,}8}{8{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{TAN}\) = Arccos (\(\dfrac{3{,}8}{8{,}7}\)) \(\approx\) 64\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{VMT}\)) = \(\dfrac{VT}{VM}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{VMT}\)) = \(\dfrac{4{,}8}{3{,}3}\)
\par\addvspace{.5em}
 On a donc VMT = ArcTan(\(\dfrac{4{,}8}{3{,}3}\)) \(\approx\) 55\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{NLR}\)) = \(\dfrac{NR}{NL}\)
\par\addvspace{.5em}
d'où  tan(20\textdegree{}) =  \(\dfrac{7}{NL}\)
\par\addvspace{.5em}
On a donc NL = \(\dfrac{7}{\tan(20^{\circ})}\) \(\approx\) 19,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{BHD}\)) = \(\dfrac{BH}{DH}\)
\par\addvspace{.5em}
  d'où    cos(45\textdegree{}) =  \(\dfrac{2{,}2}{DH}\)
\par\addvspace{.5em}
 On a donc DH =  \(\dfrac{2{,}2}{\cos(45^{\circ})}\) \(\approx\) 3,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{MSP}\)) = \(\dfrac{MS}{SP}\)
\par\addvspace{.5em}
d'où cos(77\textdegree{}) = \(\dfrac{MS}{4{,}7}\)

On a donc MS = 4,7 \(\times\) cos(77\textdegree{}) \(\approx\) 1,1 cm\end{document}
