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\begin{document}

\sloppy

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

 Dans le triangle BSM rectangle en B, on sait que :
\begin{itemize}
\item BM=10 cm
\item \(\widehat{BSM}\) = 67\textdegree{}
\end{itemize}

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

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

 Dans le triangle GDB rectangle en G, on sait que :
\begin{itemize}
\item GD=3,2 cm
\item DB=9,8 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{GDB}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle WDF rectangle en W, on sait que :
\begin{itemize}
\item WD=3,3 cm
\item DF=7,4 cm
\end{itemize}

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

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

 Dans le triangle GWD rectangle en G, on sait que :
\begin{itemize}
\item GW=3,6 cm
\item \(\widehat{GWD}\) = 61\textdegree{}
\end{itemize}

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

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

 Dans le triangle SWD rectangle en S, on sait que :
\begin{itemize}
\item SD=5,6 cm
\item \(\widehat{WDS}\) =  40\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [DW]. (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 (1.418,2.417) {67°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{BSM}\)) = \(\dfrac{BM}{BS}\)
\par\addvspace{.5em}
tan(67\textdegree{}) = \(\dfrac{10}{BS}\)
\par\addvspace{.5em}
 On a donc BS = \(\dfrac{10}{\tan(67^{\circ})}\) \(\approx\) 4,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{GDB}\)) = \(\dfrac{GD}{DB}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{GDB}\)) = \(\dfrac{3{,}2}{9{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{GDB}\) =ArcCos(\(\dfrac{3{,}2}{9{,}8}\)) \(\approx\)71\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.301,0.865) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{WFD}\)) = \(\dfrac{WD}{DF}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{WFD}\)) =\(\dfrac{3{,}3}{7{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{WFD}\) = ArcSin(\(\dfrac{3{,}3}{7{,}4}\)) \(\approx\) 26\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{GWD}\)) = \(\dfrac{GD}{GW}\)
\par\addvspace{.5em}
d'où tan(61\textdegree{}) =\(\dfrac{GD}{3{,}6}\)
\par\addvspace{.5em}
On a donc GD = 3,6 \(\times\) tan(61\textdegree{}) \(\approx\) 6,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {5,6 cm};
  \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 (4.542,0.761) {40°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{SDW}\)) = \(\dfrac{SD}{WD}\)
\par\addvspace{.5em}
  d'où    cos(40\textdegree{}) =  \(\dfrac{5{,}6}{WD}\)
\par\addvspace{.5em}
 On a donc WD =  \(\dfrac{5{,}6}{\cos(40^{\circ})}\) \(\approx\) 7,3 cm\end{document}
