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\begin{document}

\sloppy

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

 Dans le triangle GKJ rectangle en G, on sait que :
\begin{itemize}
\item GJ=4,5 cm
\item KJ=6,9 cm
\end{itemize}

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

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

 Dans le triangle PFT rectangle en P, on sait que :
\begin{itemize}
\item PF=0,6 cm
\item \(\widehat{PFT}\) = 77\textdegree{}
\end{itemize}

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

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

 Dans le triangle NDK rectangle en N, on sait que :
\begin{itemize}
\item ND=2,4 cm
\item NK=4,5 cm
\end{itemize}

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

 Dans le triangle LBS rectangle en L, on sait que :
\begin{itemize}
\item BS=4,7 cm
\item \(\widehat{BSL}\) =  38\textdegree{}
\end{itemize}

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

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

 Dans le triangle DVB rectangle en D, on sait que :
\begin{itemize}
\item DV=9,6 cm
\item \(\widehat{DVB}\) = 57\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [BV]. (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 (4.099,0.811) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{GJK}\)) = \(\dfrac{GJ}{KJ}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{GJK}\)) =  \(\dfrac{4{,}5}{6{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{GJK}\) = Arccos (\(\dfrac{4{,}5}{6{,}9}\)) \(\approx\) 49\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{PFT}\)) = \(\dfrac{PT}{PF}\)
\par\addvspace{.5em}
d'où tan(77\textdegree{}) =\(\dfrac{PT}{0{,}6}\)
\par\addvspace{.5em}
On a donc PT = 0,6 \(\times\) tan(77\textdegree{}) \(\approx\) 2,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{NDK}\)) = \(\dfrac{NK}{ND}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{NDK}\)) = \(\dfrac{4{,}5}{2{,}4}\)
\par\addvspace{.5em}
 On a donc NDK = ArcTan(\(\dfrac{4{,}5}{2{,}4}\)) \(\approx\) 62\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.662,0.750) {38°};
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 Dans le triangle LBS rectangle en L, on cherche une relation entre l'angle aigu   \(\widehat{LSB}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{LSB}\)) = \(\dfrac{LB}{BS}\)
\par\addvspace{.5em}
 D'où sin(38\textdegree{}) = \(\dfrac{LB}{4{,}7}\)

On a donc LB = 4,7 \(\times\) sin(38\textdegree{}) \(\approx\) 2,9 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) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.662,2.747) {57°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{DVB}\)) = \(\dfrac{DV}{VB}\)
\par\addvspace{.5em}
 d'où cos(57\textdegree{}) = \(\dfrac{9{,}6}{VB}\)
\par\addvspace{.5em}
On a donc VB = \(\dfrac{9{,}6}{\cos(57^{\circ})}\)\(\approx\) 17,6 cm\end{document}
