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\begin{document}

\sloppy

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

 Dans le triangle CGM rectangle en C, on sait que :
\begin{itemize}
\item CG=2,5 cm
\item CM=6 cm
\end{itemize}

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

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

 Dans le triangle HKR rectangle en H, on sait que :
\begin{itemize}
\item HR=7 cm
\item \(\widehat{HKR}\) = 62\textdegree{}
\end{itemize}

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

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

 Dans le triangle KAF rectangle en K, on sait que :
\begin{itemize}
\item KA=9,2 cm
\item \(\widehat{AFK}\) =  44\textdegree{}
\end{itemize}

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

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

 Dans le triangle HMA rectangle en H, on sait que :
\begin{itemize}
\item HM=3,1 cm
\item HA=4,4 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{HMA}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle HBV rectangle en H, on sait que :
\begin{itemize}
\item HV=7,6 cm
\item \(\widehat{HBV}\) = 78\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{CMG}\)) = \(\dfrac{CG}{CM}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{CMG}\)) =  \(\dfrac{2{,}5}{6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{CMG}\) =  ArcTan(\(\dfrac{2{,}5}{6}\)) \(\approx\) 23\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{HKR}\)) = \(\dfrac{HR}{HK}\)
\par\addvspace{.5em}
tan(62\textdegree{}) = \(\dfrac{7}{HK}\)
\par\addvspace{.5em}
 On a donc HK = \(\dfrac{7}{\tan(62^{\circ})}\) \(\approx\) 3,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{KFA}\)) = \(\dfrac{KA}{KF}\)
\par\addvspace{.5em}
d'où  tan(44\textdegree{}) =  \(\dfrac{9{,}2}{KF}\)
\par\addvspace{.5em}
On a donc KF = \(\dfrac{9{,}2}{\tan(44^{\circ})}\) \(\approx\) 9,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{HMA}\)) = \(\dfrac{HA}{HM}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{HMA}\)) = \(\dfrac{4{,}4}{3{,}1}\)
\par\addvspace{.5em}
 On a donc HMA = ArcTan(\(\dfrac{4{,}4}{3{,}1}\)) \(\approx\) 55\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.470,1.936) {78°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{HBV}\)) =  \(\dfrac{HV}{BV}\)
\par\addvspace{.5em}
 d'où sin(78\textdegree{}) = \(\dfrac{7{,}6}{BV}\)
\par\addvspace{.5em}
 On a donc BV = \(\dfrac{7{,}6}{\sin(78^{\circ})}\) \(\approx\) 7,8 cm\end{document}
