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\begin{document}

\sloppy

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

 Dans le triangle GBK rectangle en G, on sait que :
\begin{itemize}
\item GB=8,6 cm
\item \(\widehat{GBK}\) = 50\textdegree{}
\end{itemize}

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

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

 Dans le triangle HNV rectangle en H, on sait que :
\begin{itemize}
\item HN=3,3 cm
\item NV=9,6 cm
\end{itemize}

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

 Dans le triangle KDG rectangle en K, on sait que :
\begin{itemize}
\item KG=5,8 cm
\item DG=9,5 cm
\end{itemize}

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

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

 Dans le triangle STG rectangle en S, on sait que :
\begin{itemize}
\item SG=5,9 cm
\item \(\widehat{STG}\) = 51\textdegree{}
\end{itemize}

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

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

 Dans le triangle PFS rectangle en P, on sait que :
\begin{itemize}
\item PF=1,4 cm
\item \(\widehat{FSP}\) =  13\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [SF]. (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 (2.113,2.729) {50°};
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 Dans le triangle GBK rectangle en G, on cherche une relation entre l'angle aigu   \(\widehat{GBK}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{GBK}\)) = \(\dfrac{GK}{GB}\)
\par\addvspace{.5em}
d'où tan(50\textdegree{}) =\(\dfrac{GK}{8{,}6}\)
\par\addvspace{.5em}
On a donc GK = 8,6 \(\times\) tan(50\textdegree{}) \(\approx\) 10,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{HNV}\)) = \(\dfrac{HN}{NV}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{HNV}\)) = \(\dfrac{3{,}3}{9{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{HNV}\) =ArcCos(\(\dfrac{3{,}3}{9{,}6}\)) \(\approx\)70\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{KGD}\)) = \(\dfrac{KG}{DG}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{KGD}\)) =  \(\dfrac{5{,}8}{9{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{KGD}\) = Arccos (\(\dfrac{5{,}8}{9{,}5}\)) \(\approx\) 52\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{STG}\)) = \(\dfrac{SG}{ST}\)
\par\addvspace{.5em}
tan(51\textdegree{}) = \(\dfrac{5{,}9}{ST}\)
\par\addvspace{.5em}
 On a donc ST = \(\dfrac{5{,}9}{\tan(51^{\circ})}\) \(\approx\) 4,8 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 (5.288,1.443) {13°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{PSF}\)) = \(\dfrac{PF}{FS}\)
\par\addvspace{.5em}
 D'où sin(13\textdegree{}) = \(\dfrac{1{,}4}{FS}\)
\par\addvspace{.5em}
On a donc FS =\(\dfrac{1{,}4}{\sin(13^{\circ})}\) \(\approx\) 6,2 cm\end{document}
