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\begin{document}

\sloppy

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

 Dans le triangle FSB rectangle en F, on sait que :
\begin{itemize}
\item FS=8,1 cm
\item \(\widehat{FSB}\) = 46\textdegree{}
\end{itemize}

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

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

 Dans le triangle HVP rectangle en H, on sait que :
\begin{itemize}
\item HV=1,1 cm
\item HP=4,9 cm
\end{itemize}

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

 Dans le triangle RWG rectangle en R, on sait que :
\begin{itemize}
\item RG=2,8 cm
\item \(\widehat{RWG}\) = 70\textdegree{}
\end{itemize}

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

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

 Dans le triangle TBS rectangle en T, on sait que :
\begin{itemize}
\item TB=2,4 cm
\item TS=6,6 cm
\end{itemize}

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

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

 Dans le triangle KWT rectangle en K, on sait que :
\begin{itemize}
\item KT=3,7 cm
\item \(\widehat{WTK}\) =  40\textdegree{}
\end{itemize}

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

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

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

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{FSB}\)) = \(\dfrac{FB}{FS}\)
\par\addvspace{.5em}
d'où tan(46\textdegree{}) =\(\dfrac{FB}{8{,}1}\)
\par\addvspace{.5em}
On a donc FB = 8,1 \(\times\) tan(46\textdegree{}) \(\approx\) 8,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{HVP}\)) = \(\dfrac{HP}{HV}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{HVP}\)) = \(\dfrac{4{,}9}{1{,}1}\)
\par\addvspace{.5em}
 On a donc HVP = ArcTan(\(\dfrac{4{,}9}{1{,}1}\)) \(\approx\) 77\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{RWG}\)) = \(\dfrac{RG}{RW}\)
\par\addvspace{.5em}
tan(70\textdegree{}) = \(\dfrac{2{,}8}{RW}\)
\par\addvspace{.5em}
 On a donc RW = \(\dfrac{2{,}8}{\tan(70^{\circ})}\) \(\approx\) 1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{TSB}\)) = \(\dfrac{TB}{TS}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{TSB}\)) =  \(\dfrac{2{,}4}{6{,}6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{TSB}\) =  ArcTan(\(\dfrac{2{,}4}{6{,}6}\)) \(\approx\) 20\textdegree{}
\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 (4.542,0.761) {40°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{KTW}\)) = \(\dfrac{KT}{WT}\)
\par\addvspace{.5em}
  d'où    cos(40\textdegree{}) =  \(\dfrac{3{,}7}{WT}\)
\par\addvspace{.5em}
 On a donc WT =  \(\dfrac{3{,}7}{\cos(40^{\circ})}\) \(\approx\) 4,8 cm\end{document}
