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\begin{document}

\sloppy

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

 Dans le triangle GFH rectangle en G, on sait que :
\begin{itemize}
\item GH=3,5 cm
\item FH=10 cm
\end{itemize}

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

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

 Dans le triangle AJW rectangle en A, on sait que :
\begin{itemize}
\item AW=3,8 cm
\item JW=7,3 cm
\end{itemize}

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

 Dans le triangle SBW rectangle en S, on sait que :
\begin{itemize}
\item SW=2,9 cm
\item \(\widehat{BWS}\) =  35\textdegree{}
\end{itemize}

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

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

 Dans le triangle WLH rectangle en W, on sait que :
\begin{itemize}
\item LH=7 cm
\item \(\widehat{WLH}\) = 49\textdegree{}
\end{itemize}

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

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

 Dans le triangle LCR rectangle en L, on sait que :
\begin{itemize}
\item LC=2,7 cm
\item \(\widehat{CRL}\) =  36\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [LR]. (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 (3.475,0.913) {?};
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 Dans le triangle GFH rectangle en G, on cherche une relation entre l'angle aigu   \(\widehat{GHF}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{GHF}\)) = \(\dfrac{GH}{FH}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{GHF}\)) =  \(\dfrac{3{,}5}{10}\)
\par\addvspace{.5em}
On a donc \(\widehat{GHF}\) = Arccos (\(\dfrac{3{,}5}{10}\)) \(\approx\) 70\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{AJW}\)) =  \(\dfrac{AW}{JW}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{AJW}\)) = \(\dfrac{3{,}8}{7{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{AJW}\) =  ArcSin(\(\dfrac{3{,}8}{7{,}3}\)) \(\approx\) 31\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

cos(\(\widehat{SWB}\)) = \(\dfrac{SW}{BW}\)
\par\addvspace{.5em}
  d'où    cos(35\textdegree{}) =  \(\dfrac{2{,}9}{BW}\)
\par\addvspace{.5em}
 On a donc BW =  \(\dfrac{2{,}9}{\cos(35^{\circ})}\) \(\approx\) 3,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.166,2.727) {49°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{WLH}\)) = \(\dfrac{WL}{LH}\)
\par\addvspace{.5em}
d'où cos(49\textdegree{}) = \(\dfrac{WL}{7}\)

On a donc WL = 7 \(\times\) cos(49\textdegree{}) \(\approx\) 4,6 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) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.793,0.739) {36°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{LRC}\)) = \(\dfrac{LC}{LR}\)
\par\addvspace{.5em}
d'où  tan(36\textdegree{}) =  \(\dfrac{2{,}7}{LR}\)
\par\addvspace{.5em}
On a donc LR = \(\dfrac{2{,}7}{\tan(36^{\circ})}\) \(\approx\) 3,7 cm\end{document}
