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\begin{document}

\sloppy

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

 Dans le triangle VWR rectangle en V, on sait que :
\begin{itemize}
\item VR=4,8 cm
\item \(\widehat{WRV}\) =  22\textdegree{}
\end{itemize}

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

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

 Dans le triangle LSH rectangle en L, on sait que :
\begin{itemize}
\item LS=3 cm
\item LH=5,2 cm
\end{itemize}

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

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

 Dans le triangle HLB rectangle en H, on sait que :
\begin{itemize}
\item LB=5,8 cm
\item \(\widehat{LBH}\) =  16\textdegree{}
\end{itemize}

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

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

 Dans le triangle GSH rectangle en G, on sait que :
\begin{itemize}
\item GH=5,8 cm
\item SH=9,9 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{GSH}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle BAD rectangle en B, on sait que :
\begin{itemize}
\item BD=9,9 cm
\item \(\widehat{ADB}\) =  15\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{VRW}\)) = \(\dfrac{VR}{WR}\)
\par\addvspace{.5em}
  d'où    cos(22\textdegree{}) =  \(\dfrac{4{,}8}{WR}\)
\par\addvspace{.5em}
 On a donc WR =  \(\dfrac{4{,}8}{\cos(22^{\circ})}\) \(\approx\) 5,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{LHS}\)) = \(\dfrac{LS}{LH}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{LHS}\)) =  \(\dfrac{3}{5{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{LHS}\) =  ArcTan(\(\dfrac{3}{5{,}2}\)) \(\approx\) 30\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{HBL}\)) = \(\dfrac{HB}{LB}\)
\par\addvspace{.5em}
  d'où    cos(16\textdegree{}) =  \(\dfrac{HB}{5{,}8}\)

On a donc HB = 5,8 \(\times\) cos(16\textdegree{}) \(\approx\) 5{,}6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{GSH}\)) =  \(\dfrac{GH}{SH}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{GSH}\)) = \(\dfrac{5{,}8}{9{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{GSH}\) =  ArcSin(\(\dfrac{5{,}8}{9{,}9}\)) \(\approx\) 36\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.289,1.364) {15°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{BDA}\)) = \(\dfrac{BA}{BD}\)
\par\addvspace{.5em}
d'où  tan(15\textdegree{}) =  \(\dfrac{BA}{9{,}9}\)
\par\addvspace{.5em}
On a donc BA =  9,9 \(\times\) tan(15\textdegree{})  \(\approx\) 2,7 cm\end{document}
