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\begin{document}

\sloppy

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

 Dans le triangle LCV rectangle en L, on sait que :
\begin{itemize}
\item LV=7,9 cm
\item \(\widehat{LCV}\) = 53\textdegree{}
\end{itemize}

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

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

 Dans le triangle PHT rectangle en P, on sait que :
\begin{itemize}
\item PH=0,7 cm
\item \(\widehat{HTP}\) =  15\textdegree{}
\end{itemize}

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

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

 Dans le triangle MGN rectangle en M, on sait que :
\begin{itemize}
\item MG=3,1 cm
\item GN=8 cm
\end{itemize}

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

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

 Dans le triangle DMP rectangle en D, on sait que :
\begin{itemize}
\item DP=10 cm
\item \(\widehat{DMP}\) = 55\textdegree{}
\end{itemize}

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

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

 Dans le triangle DRF rectangle en D, on sait que :
\begin{itemize}
\item DF=5,9 cm
\item RF=9,4 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{DRF}\). \newpage
\section*{\textcolor{red}{Correction}}

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

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

\par\addvspace{.5em}
tan(\(\widehat{LCV}\)) = \(\dfrac{LV}{LC}\)
\par\addvspace{.5em}
tan(53\textdegree{}) = \(\dfrac{7{,}9}{LC}\)
\par\addvspace{.5em}
 On a donc LC = \(\dfrac{7{,}9}{\tan(53^{\circ})}\) \(\approx\) 6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{PTH}\)) = \(\dfrac{PH}{PT}\)
\par\addvspace{.5em}
d'où  tan(15\textdegree{}) =  \(\dfrac{0{,}7}{PT}\)
\par\addvspace{.5em}
On a donc PT = \(\dfrac{0{,}7}{\tan(15^{\circ})}\) \(\approx\) 2,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{MNG}\)) = \(\dfrac{MG}{GN}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{MNG}\)) =\(\dfrac{3{,}1}{8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{MNG}\) = ArcSin(\(\dfrac{3{,}1}{8}\)) \(\approx\) 23\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{DMP}\)) =  \(\dfrac{DP}{MP}\)
\par\addvspace{.5em}
 d'où sin(55\textdegree{}) = \(\dfrac{10}{MP}\)
\par\addvspace{.5em}
 On a donc MP = \(\dfrac{10}{\sin(55^{\circ})}\) \(\approx\) 12,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.593,2.705) {?};
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 Dans le triangle DRF rectangle en D, on cherche une relation entre l'angle aigu   \(\widehat{DRF}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{DRF}\)) =  \(\dfrac{DF}{RF}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{DRF}\)) = \(\dfrac{5{,}9}{9{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{DRF}\) =  ArcSin(\(\dfrac{5{,}9}{9{,}4}\)) \(\approx\) 39\textdegree{}\end{document}
