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\begin{document}

\sloppy

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

 Dans le triangle TMA rectangle en T, on sait que :
\begin{itemize}
\item TA=1,8 cm
\item \(\widehat{TMA}\) = 47\textdegree{}
\end{itemize}

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

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

 Dans le triangle MPZ rectangle en M, on sait que :
\begin{itemize}
\item PZ=3,5 cm
\item \(\widehat{MPZ}\) = 78\textdegree{}
\end{itemize}

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

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

 Dans le triangle TGK rectangle en T, on sait que :
\begin{itemize}
\item TG=3,3 cm
\item GK=9,4 cm
\end{itemize}

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

 Dans le triangle KVF rectangle en K, on sait que :
\begin{itemize}
\item KV=5,4 cm
\item \(\widehat{VFK}\) =  34\textdegree{}
\end{itemize}

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

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

 Dans le triangle CHN rectangle en C, on sait que :
\begin{itemize}
\item CH=3,2 cm
\item CN=6,4 cm
\end{itemize}

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

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

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

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 Dans le triangle TMA rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TMA}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{TMA}\)) =  \(\dfrac{TA}{MA}\)
\par\addvspace{.5em}
 d'où sin(47\textdegree{}) = \(\dfrac{1{,}8}{MA}\)
\par\addvspace{.5em}
 On a donc MA = \(\dfrac{1{,}8}{\sin(47^{\circ})}\) \(\approx\) 2,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{MPZ}\)) = \(\dfrac{MP}{PZ}\)
\par\addvspace{.5em}
d'où cos(78\textdegree{}) = \(\dfrac{MP}{3{,}5}\)

On a donc MP = 3,5 \(\times\) cos(78\textdegree{}) \(\approx\) 0,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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 Dans le triangle TGK rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TGK}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{TGK}\)) = \(\dfrac{TG}{GK}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{TGK}\)) = \(\dfrac{3{,}3}{9{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{TGK}\) =ArcCos(\(\dfrac{3{,}3}{9{,}4}\)) \(\approx\)69\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{KFV}\)) = \(\dfrac{KV}{KF}\)
\par\addvspace{.5em}
d'où  tan(34\textdegree{}) =  \(\dfrac{5{,}4}{KF}\)
\par\addvspace{.5em}
On a donc KF = \(\dfrac{5{,}4}{\tan(34^{\circ})}\) \(\approx\) 8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.301,0.862) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{CNH}\)) = \(\dfrac{CH}{CN}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{CNH}\)) =  \(\dfrac{3{,}2}{6{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{CNH}\) =  ArcTan(\(\dfrac{3{,}2}{6{,}4}\)) \(\approx\) 27\textdegree{}\end{document}
