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\begin{document}

\sloppy

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

 Dans le triangle MTH rectangle en M, on sait que :
\begin{itemize}
\item TH=0,8 cm
\item \(\widehat{MTH}\) = 46\textdegree{}
\end{itemize}

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

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

 Dans le triangle RGB rectangle en R, on sait que :
\begin{itemize}
\item RB=3,7 cm
\item GB=7,5 cm
\end{itemize}

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

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

 Dans le triangle RCB rectangle en R, on sait que :
\begin{itemize}
\item RC=2,3 cm
\item RB=5,9 cm
\end{itemize}

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

 Dans le triangle HKN rectangle en H, on sait que :
\begin{itemize}
\item HN=9 cm
\item \(\widehat{KNH}\) =  37\textdegree{}
\end{itemize}

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

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

 Dans le triangle LFJ rectangle en L, on sait que :
\begin{itemize}
\item FJ=7 cm
\item \(\widehat{LFJ}\) = 73\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [LJ]. (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 (3.667,2.000) {0,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.311,2.720) {46°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{MTH}\)) = \(\dfrac{MT}{TH}\)
\par\addvspace{.5em}
d'où cos(46\textdegree{}) = \(\dfrac{MT}{0{,}8}\)

On a donc MT = 0,8 \(\times\) cos(46\textdegree{}) \(\approx\) 0,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{RBG}\)) = \(\dfrac{RB}{GB}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{RBG}\)) =  \(\dfrac{3{,}7}{7{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{RBG}\) = Arccos (\(\dfrac{3{,}7}{7{,}5}\)) \(\approx\) 60\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{RCB}\)) = \(\dfrac{RB}{RC}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{RCB}\)) = \(\dfrac{5{,}9}{2{,}3}\)
\par\addvspace{.5em}
 On a donc RCB = ArcTan(\(\dfrac{5{,}9}{2{,}3}\)) \(\approx\) 69\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{HNK}\)) = \(\dfrac{HN}{KN}\)
\par\addvspace{.5em}
  d'où    cos(37\textdegree{}) =  \(\dfrac{9}{KN}\)
\par\addvspace{.5em}
 On a donc KN =  \(\dfrac{9}{\cos(37^{\circ})}\) \(\approx\) 11,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.918) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.447,2.146) {73°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{LFJ}\)) =  \(\dfrac{LJ}{FJ}\)
\par\addvspace{.5em}
 d'où sin(73\textdegree{}) = \(\dfrac{LJ}{7}\)

On a donc LJ = 7 \(\times\) sin(73\textdegree{}) \(\approx\) 6,7 cm\end{document}
