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\begin{document}

\sloppy

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

 Dans le triangle GHR rectangle en G, on sait que :
\begin{itemize}
\item GH=1 cm
\item HR=9,6 cm
\end{itemize}

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

 Dans le triangle TVS rectangle en T, on sait que :
\begin{itemize}
\item TV=1,2 cm
\item TS=4,5 cm
\end{itemize}

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

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

 Dans le triangle AKJ rectangle en A, on sait que :
\begin{itemize}
\item AJ=1,3 cm
\item \(\widehat{AKJ}\) = 62\textdegree{}
\end{itemize}

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

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

 Dans le triangle GNW rectangle en G, on sait que :
\begin{itemize}
\item GW=2,7 cm
\item \(\widehat{GNW}\) = 65\textdegree{}
\end{itemize}

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

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

 Dans le triangle CLR rectangle en C, on sait que :
\begin{itemize}
\item LR=4,1 cm
\item \(\widehat{CLR}\) = 48\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{GHR}\)) = \(\dfrac{GH}{HR}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{GHR}\)) = \(\dfrac{1}{9{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{GHR}\) =ArcCos(\(\dfrac{1}{9{,}6}\)) \(\approx\)84\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{TSV}\)) = \(\dfrac{TV}{TS}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{TSV}\)) =  \(\dfrac{1{,}2}{4{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{TSV}\) =  ArcTan(\(\dfrac{1{,}2}{4{,}5}\)) \(\approx\) 15\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{AKJ}\)) = \(\dfrac{AJ}{AK}\)
\par\addvspace{.5em}
tan(62\textdegree{}) = \(\dfrac{1{,}3}{AK}\)
\par\addvspace{.5em}
 On a donc AK = \(\dfrac{1{,}3}{\tan(62^{\circ})}\) \(\approx\) 0,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{GNW}\)) =  \(\dfrac{GW}{NW}\)
\par\addvspace{.5em}
 d'où sin(65\textdegree{}) = \(\dfrac{2{,}7}{NW}\)
\par\addvspace{.5em}
 On a donc NW = \(\dfrac{2{,}7}{\sin(65^{\circ})}\) \(\approx\) 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) {4,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.216,2.724) {48°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{CLR}\)) =  \(\dfrac{CR}{LR}\)
\par\addvspace{.5em}
 d'où sin(48\textdegree{}) = \(\dfrac{CR}{4{,}1}\)

On a donc CR = 4,1 \(\times\) sin(48\textdegree{}) \(\approx\) 3 cm\end{document}
