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\begin{document}

\sloppy

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

 Dans le triangle TCS rectangle en T, on sait que :
\begin{itemize}
\item CS=4,4 cm
\item \(\widehat{CST}\) =  36\textdegree{}
\end{itemize}

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

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

 Dans le triangle PVB rectangle en P, on sait que :
\begin{itemize}
\item PB=6,1 cm
\item VB=7,3 cm
\end{itemize}

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

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

 Dans le triangle BWC rectangle en B, on sait que :
\begin{itemize}
\item BW=2 cm
\item \(\widehat{WCB}\) =  35\textdegree{}
\end{itemize}

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

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

 Dans le triangle DSV rectangle en D, on sait que :
\begin{itemize}
\item DS=2,3 cm
\item \(\widehat{DSV}\) = 61\textdegree{}
\end{itemize}

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

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

 Dans le triangle BHL rectangle en B, on sait que :
\begin{itemize}
\item BH=3,3 cm
\item BL=6,5 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{BHL}\). \newpage
\section*{\textcolor{red}{Correction}}

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

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

\par\addvspace{.5em}
sin(\(\widehat{TSC}\)) = \(\dfrac{TC}{CS}\)
\par\addvspace{.5em}
 D'où sin(36\textdegree{}) = \(\dfrac{TC}{4{,}4}\)

On a donc TC = 4,4 \(\times\) sin(36\textdegree{}) \(\approx\) 2,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{PBV}\)) = \(\dfrac{PB}{VB}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{PBV}\)) =  \(\dfrac{6{,}1}{7{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{PBV}\) = Arccos (\(\dfrac{6{,}1}{7{,}3}\)) \(\approx\) 33\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{BCW}\)) = \(\dfrac{BW}{WC}\)
\par\addvspace{.5em}
 D'où sin(35\textdegree{}) = \(\dfrac{2}{WC}\)
\par\addvspace{.5em}
On a donc WC =\(\dfrac{2}{\sin(35^{\circ})}\) \(\approx\) 3,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{DSV}\)) = \(\dfrac{DV}{DS}\)
\par\addvspace{.5em}
d'où tan(61\textdegree{}) =\(\dfrac{DV}{2{,}3}\)
\par\addvspace{.5em}
On a donc DV = 2,3 \(\times\) tan(61\textdegree{}) \(\approx\) 4,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{BHL}\)) = \(\dfrac{BL}{BH}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{BHL}\)) = \(\dfrac{6{,}5}{3{,}3}\)
\par\addvspace{.5em}
 On a donc BHL = ArcTan(\(\dfrac{6{,}5}{3{,}3}\)) \(\approx\) 63\textdegree{}\end{document}
