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\begin{document}

\sloppy

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

 Dans le triangle HKL rectangle en H, on sait que :
\begin{itemize}
\item HL=7,7 cm
\item \(\widehat{HKL}\) = 57\textdegree{}
\end{itemize}

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

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

 Dans le triangle VHC rectangle en V, on sait que :
\begin{itemize}
\item HC=6,9 cm
\item \(\widehat{VHC}\) = 72\textdegree{}
\end{itemize}

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

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

 Dans le triangle TFS rectangle en T, on sait que :
\begin{itemize}
\item TF=5,1 cm
\item \(\widehat{FST}\) =  28\textdegree{}
\end{itemize}

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

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

 Dans le triangle SWL rectangle en S, on sait que :
\begin{itemize}
\item SW=1,7 cm
\item SL=5,2 cm
\end{itemize}

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

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

 Dans le triangle SMA rectangle en S, on sait que :
\begin{itemize}
\item SA=4,1 cm
\item MA=9,2 cm
\end{itemize}

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{HKL}\)) =  \(\dfrac{HL}{KL}\)
\par\addvspace{.5em}
 d'où sin(57\textdegree{}) = \(\dfrac{7{,}7}{KL}\)
\par\addvspace{.5em}
 On a donc KL = \(\dfrac{7{,}7}{\sin(57^{\circ})}\) \(\approx\) 9,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{VHC}\)) = \(\dfrac{VH}{HC}\)
\par\addvspace{.5em}
d'où cos(72\textdegree{}) = \(\dfrac{VH}{6{,}9}\)

On a donc VH = 6,9 \(\times\) cos(72\textdegree{}) \(\approx\) 2,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{TSF}\)) = \(\dfrac{TF}{TS}\)
\par\addvspace{.5em}
d'où  tan(28\textdegree{}) =  \(\dfrac{5{,}1}{TS}\)
\par\addvspace{.5em}
On a donc TS = \(\dfrac{5{,}1}{\tan(28^{\circ})}\) \(\approx\) 9,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{SLW}\)) = \(\dfrac{SW}{SL}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{SLW}\)) =  \(\dfrac{1{,}7}{5{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{SLW}\) =  ArcTan(\(\dfrac{1{,}7}{5{,}2}\)) \(\approx\) 18\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.956,2.684) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{SMA}\)) =  \(\dfrac{SA}{MA}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{SMA}\)) = \(\dfrac{4{,}1}{9{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{SMA}\) =  ArcSin(\(\dfrac{4{,}1}{9{,}2}\)) \(\approx\) 26\textdegree{}\end{document}
