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\begin{document}

\sloppy

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

 Dans le triangle GHW rectangle en G, on sait que :
\begin{itemize}
\item GW=6,1 cm
\item HW=9,1 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{GHW}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle MDW rectangle en M, on sait que :
\begin{itemize}
\item MD=6,4 cm
\item \(\widehat{DWM}\) =  23\textdegree{}
\end{itemize}

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

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

 Dans le triangle TJP rectangle en T, on sait que :
\begin{itemize}
\item TP=3,8 cm
\item JP=9,3 cm
\end{itemize}

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

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

 Dans le triangle KTG rectangle en K, on sait que :
\begin{itemize}
\item KT=7 cm
\item \(\widehat{TGK}\) =  10\textdegree{}
\end{itemize}

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

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

 Dans le triangle VTS rectangle en V, on sait que :
\begin{itemize}
\item VS=1,8 cm
\item \(\widehat{VTS}\) = 46\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [VT]. (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 (2.475,2.711) {?};
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 Dans le triangle GHW rectangle en G, on cherche une relation entre l'angle aigu   \(\widehat{GHW}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{GHW}\)) =  \(\dfrac{GW}{HW}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{GHW}\)) = \(\dfrac{6{,}1}{9{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{GHW}\) =  ArcSin(\(\dfrac{6{,}1}{9{,}1}\)) \(\approx\) 42\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{MWD}\)) = \(\dfrac{MD}{DW}\)
\par\addvspace{.5em}
 D'où sin(23\textdegree{}) = \(\dfrac{6{,}4}{DW}\)
\par\addvspace{.5em}
On a donc DW =\(\dfrac{6{,}4}{\sin(23^{\circ})}\) \(\approx\) 16,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{TPJ}\)) = \(\dfrac{TP}{JP}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{TPJ}\)) =  \(\dfrac{3{,}8}{9{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{TPJ}\) = Arccos (\(\dfrac{3{,}8}{9{,}3}\)) \(\approx\) 66\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{KGT}\)) = \(\dfrac{KT}{KG}\)
\par\addvspace{.5em}
d'où  tan(10\textdegree{}) =  \(\dfrac{7}{KG}\)
\par\addvspace{.5em}
On a donc KG = \(\dfrac{7}{\tan(10^{\circ})}\) \(\approx\) 39,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.311,2.720) {46°};
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 Dans le triangle VTS rectangle en V, on cherche une relation entre l'angle aigu   \(\widehat{VTS}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{VTS}\)) = \(\dfrac{VS}{VT}\)
\par\addvspace{.5em}
tan(46\textdegree{}) = \(\dfrac{1{,}8}{VT}\)
\par\addvspace{.5em}
 On a donc VT = \(\dfrac{1{,}8}{\tan(46^{\circ})}\) \(\approx\) 1,7 cm\end{document}
