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\begin{document}

\sloppy

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

 Dans le triangle BFL rectangle en B, on sait que :
\begin{itemize}
\item FL=9,2 cm
\item \(\widehat{BFL}\) = 47\textdegree{}
\end{itemize}

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

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

 Dans le triangle TBA rectangle en T, on sait que :
\begin{itemize}
\item TA=5,1 cm
\item \(\widehat{TBA}\) = 54\textdegree{}
\end{itemize}

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

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

 Dans le triangle HTV rectangle en H, on sait que :
\begin{itemize}
\item HV=5,1 cm
\item TV=7,9 cm
\end{itemize}

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

 Dans le triangle RSW rectangle en R, on sait que :
\begin{itemize}
\item RW=5,2 cm
\item \(\widehat{SWR}\) =  45\textdegree{}
\end{itemize}

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

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

 Dans le triangle JWK rectangle en J, on sait que :
\begin{itemize}
\item JW=2,5 cm
\item WK=7 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{BFL}\)) =  \(\dfrac{BL}{FL}\)
\par\addvspace{.5em}
 d'où sin(47\textdegree{}) = \(\dfrac{BL}{9{,}2}\)

On a donc BL = 9,2 \(\times\) sin(47\textdegree{}) \(\approx\) 6,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{TBA}\)) = \(\dfrac{TA}{TB}\)
\par\addvspace{.5em}
tan(54\textdegree{}) = \(\dfrac{5{,}1}{TB}\)
\par\addvspace{.5em}
 On a donc TB = \(\dfrac{5{,}1}{\tan(54^{\circ})}\) \(\approx\) 3,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{HTV}\)) =  \(\dfrac{HV}{TV}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{HTV}\)) = \(\dfrac{5{,}1}{7{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{HTV}\) =  ArcSin(\(\dfrac{5{,}1}{7{,}9}\)) \(\approx\) 40\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{RWS}\)) = \(\dfrac{RW}{SW}\)
\par\addvspace{.5em}
  d'où    cos(45\textdegree{}) =  \(\dfrac{5{,}2}{SW}\)
\par\addvspace{.5em}
 On a donc SW =  \(\dfrac{5{,}2}{\cos(45^{\circ})}\) \(\approx\) 7,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{JKW}\)) = \(\dfrac{JW}{WK}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{JKW}\)) =\(\dfrac{2{,}5}{7}\)
\par\addvspace{.5em}
 On a donc \(\widehat{JKW}\) = ArcSin(\(\dfrac{2{,}5}{7}\)) \(\approx\) 21\textdegree{}\end{document}
