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\begin{document}

\sloppy

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

 Dans le triangle TPB rectangle en T, on sait que :
\begin{itemize}
\item TP=3,8 cm
\item \(\widehat{PBT}\) =  40\textdegree{}
\end{itemize}

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

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

 Dans le triangle NGT rectangle en N, on sait que :
\begin{itemize}
\item NG=3,3 cm
\item GT=8,6 cm
\end{itemize}

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

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

 Dans le triangle LDG rectangle en L, on sait que :
\begin{itemize}
\item DG=4,2 cm
\item \(\widehat{DGL}\) =  13\textdegree{}
\end{itemize}

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

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

 Dans le triangle SGJ rectangle en S, on sait que :
\begin{itemize}
\item SG=7,4 cm
\item \(\widehat{GJS}\) =  44\textdegree{}
\end{itemize}

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

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

 Dans le triangle SMG rectangle en S, on sait que :
\begin{itemize}
\item SM=1,4 cm
\item MG=7 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{SMG}\). \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 (4.542,0.761) {40°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{TBP}\)) = \(\dfrac{TP}{TB}\)
\par\addvspace{.5em}
d'où  tan(40\textdegree{}) =  \(\dfrac{3{,}8}{TB}\)
\par\addvspace{.5em}
On a donc TB = \(\dfrac{3{,}8}{\tan(40^{\circ})}\) \(\approx\) 4,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{NTG}\)) = \(\dfrac{NG}{GT}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{NTG}\)) =\(\dfrac{3{,}3}{8{,}6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{NTG}\) = ArcSin(\(\dfrac{3{,}3}{8{,}6}\)) \(\approx\) 23\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{LGD}\)) = \(\dfrac{LD}{DG}\)
\par\addvspace{.5em}
 D'où sin(13\textdegree{}) = \(\dfrac{LD}{4{,}2}\)

On a donc LD = 4,2 \(\times\) sin(13\textdegree{}) \(\approx\) 0,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{SJG}\)) = \(\dfrac{SG}{GJ}\)
\par\addvspace{.5em}
 D'où sin(44\textdegree{}) = \(\dfrac{7{,}4}{GJ}\)
\par\addvspace{.5em}
On a donc GJ =\(\dfrac{7{,}4}{\sin(44^{\circ})}\) \(\approx\) 10,7 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) {7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.472,1.917) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{SMG}\)) = \(\dfrac{SM}{MG}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{SMG}\)) = \(\dfrac{1{,}4}{7}\)
\par\addvspace{.5em}
On a donc \(\widehat{SMG}\) =ArcCos(\(\dfrac{1{,}4}{7}\)) \(\approx\)78\textdegree{}\end{document}
