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\begin{document}

\sloppy

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

 Dans le triangle MGA rectangle en M, on sait que :
\begin{itemize}
\item MA=4,2 cm
\item \(\widehat{MGA}\) = 64\textdegree{}
\end{itemize}

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

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

 Dans le triangle KNV rectangle en K, on sait que :
\begin{itemize}
\item KN=3,2 cm
\item KV=5,4 cm
\end{itemize}

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

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

 Dans le triangle TKV rectangle en T, on sait que :
\begin{itemize}
\item TV=2,2 cm
\item \(\widehat{TKV}\) = 55\textdegree{}
\end{itemize}

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

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

 Dans le triangle DKT rectangle en D, on sait que :
\begin{itemize}
\item DK=0,4 cm
\item \(\widehat{KTD}\) =  10\textdegree{}
\end{itemize}

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

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

 Dans le triangle BPT rectangle en B, on sait que :
\begin{itemize}
\item BP=1,3 cm
\item PT=7,4 cm
\end{itemize}

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{MGA}\)) = \(\dfrac{MA}{MG}\)
\par\addvspace{.5em}
tan(64\textdegree{}) = \(\dfrac{4{,}2}{MG}\)
\par\addvspace{.5em}
 On a donc MG = \(\dfrac{4{,}2}{\tan(64^{\circ})}\) \(\approx\) 2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{KVN}\)) = \(\dfrac{KN}{KV}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{KVN}\)) =  \(\dfrac{3{,}2}{5{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{KVN}\) =  ArcTan(\(\dfrac{3{,}2}{5{,}4}\)) \(\approx\) 31\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{TKV}\)) =  \(\dfrac{TV}{KV}\)
\par\addvspace{.5em}
 d'où sin(55\textdegree{}) = \(\dfrac{2{,}2}{KV}\)
\par\addvspace{.5em}
 On a donc KV = \(\dfrac{2{,}2}{\sin(55^{\circ})}\) \(\approx\) 2,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{DTK}\)) = \(\dfrac{DK}{DT}\)
\par\addvspace{.5em}
d'où  tan(10\textdegree{}) =  \(\dfrac{0{,}4}{DT}\)
\par\addvspace{.5em}
On a donc DT = \(\dfrac{0{,}4}{\tan(10^{\circ})}\) \(\approx\) 2,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {7,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.478,1.859) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{BPT}\)) = \(\dfrac{BP}{PT}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{BPT}\)) = \(\dfrac{1{,}3}{7{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{BPT}\) =ArcCos(\(\dfrac{1{,}3}{7{,}4}\)) \(\approx\)80\textdegree{}\end{document}
