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\begin{document}

\sloppy

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

 Dans le triangle ASV rectangle en A, on sait que :
\begin{itemize}
\item AV=3,8 cm
\item \(\widehat{SVA}\) =  37\textdegree{}
\end{itemize}

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

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

 Dans le triangle WKM rectangle en W, on sait que :
\begin{itemize}
\item WK=3 cm
\item KM=7,7 cm
\end{itemize}

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

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

 Dans le triangle PWB rectangle en P, on sait que :
\begin{itemize}
\item WB=8,7 cm
\item \(\widehat{PWB}\) = 61\textdegree{}
\end{itemize}

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

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

 Dans le triangle WLK rectangle en W, on sait que :
\begin{itemize}
\item LK=9,3 cm
\item \(\widehat{LKW}\) =  10\textdegree{}
\end{itemize}

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

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

 Dans le triangle LVA rectangle en L, on sait que :
\begin{itemize}
\item LV=2,5 cm
\item VA=9,1 cm
\end{itemize}

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

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

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.726,0.745) {37°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{AVS}\)) = \(\dfrac{AV}{SV}\)
\par\addvspace{.5em}
  d'où    cos(37\textdegree{}) =  \(\dfrac{3{,}8}{SV}\)
\par\addvspace{.5em}
 On a donc SV =  \(\dfrac{3{,}8}{\cos(37^{\circ})}\) \(\approx\) 4,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{WMK}\)) = \(\dfrac{WK}{KM}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{WMK}\)) =\(\dfrac{3}{7{,}7}\)
\par\addvspace{.5em}
 On a donc \(\widehat{WMK}\) = ArcSin(\(\dfrac{3}{7{,}7}\)) \(\approx\) 23\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{PWB}\)) = \(\dfrac{PW}{WB}\)
\par\addvspace{.5em}
d'où cos(61\textdegree{}) = \(\dfrac{PW}{8{,}7}\)

On a donc PW = 8,7 \(\times\) cos(61\textdegree{}) \(\approx\) 4,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{WKL}\)) = \(\dfrac{WK}{LK}\)
\par\addvspace{.5em}
  d'où    cos(10\textdegree{}) =  \(\dfrac{WK}{9{,}3}\)

On a donc WK = 9,3 \(\times\) cos(10\textdegree{}) \(\approx\) 9{,}2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.451,2.101) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{LVA}\)) = \(\dfrac{LV}{VA}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{LVA}\)) = \(\dfrac{2{,}5}{9{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{LVA}\) =ArcCos(\(\dfrac{2{,}5}{9{,}1}\)) \(\approx\)74\textdegree{}\end{document}
