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\begin{document}

\sloppy

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

 Dans le triangle GBK rectangle en G, on sait que :
\begin{itemize}
\item GB=0,6 cm
\item \(\widehat{GBK}\) = 59\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZCN rectangle en Z, on sait que :
\begin{itemize}
\item CN=9 cm
\item \(\widehat{CNZ}\) =  42\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZBR rectangle en Z, on sait que :
\begin{itemize}
\item ZR=1,5 cm
\item \(\widehat{ZBR}\) = 60\textdegree{}
\end{itemize}

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

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

 Dans le triangle MAT rectangle en M, on sait que :
\begin{itemize}
\item MA=3,2 cm
\item AT=7,5 cm
\end{itemize}

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

 Dans le triangle MWV rectangle en M, on sait que :
\begin{itemize}
\item MV=4 cm
\item WV=7,5 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{GBK}\)) = \(\dfrac{GK}{GB}\)
\par\addvspace{.5em}
d'où tan(59\textdegree{}) =\(\dfrac{GK}{0{,}6}\)
\par\addvspace{.5em}
On a donc GK = 0,6 \(\times\) tan(59\textdegree{}) \(\approx\) 1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{ZNC}\)) = \(\dfrac{ZC}{CN}\)
\par\addvspace{.5em}
 D'où sin(42\textdegree{}) = \(\dfrac{ZC}{9}\)

On a donc ZC = 9 \(\times\) sin(42\textdegree{}) \(\approx\) 6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{ZBR}\)) =  \(\dfrac{ZR}{BR}\)
\par\addvspace{.5em}
 d'où sin(60\textdegree{}) = \(\dfrac{1{,}5}{BR}\)
\par\addvspace{.5em}
 On a donc BR = \(\dfrac{1{,}5}{\sin(60^{\circ})}\) \(\approx\) 1,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{MAT}\)) = \(\dfrac{MA}{AT}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{MAT}\)) = \(\dfrac{3{,}2}{7{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{MAT}\) =ArcCos(\(\dfrac{3{,}2}{7{,}5}\)) \(\approx\)65\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.799,0.855) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{MVW}\)) = \(\dfrac{MV}{WV}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{MVW}\)) =  \(\dfrac{4}{7{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{MVW}\) = Arccos (\(\dfrac{4}{7{,}5}\)) \(\approx\) 58\textdegree{} \end{document}
