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\begin{document}

\sloppy

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

 Dans le triangle WVS rectangle en W, on sait que :
\begin{itemize}
\item WV=2,2 cm
\item WS=5,8 cm
\end{itemize}

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

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

 Dans le triangle ZGJ rectangle en Z, on sait que :
\begin{itemize}
\item ZJ=5,9 cm
\item GJ=7,2 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{ZGJ}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle KST rectangle en K, on sait que :
\begin{itemize}
\item KS=7,5 cm
\item \(\widehat{KST}\) = 64\textdegree{}
\end{itemize}

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

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

 Dans le triangle FSR rectangle en F, on sait que :
\begin{itemize}
\item FR=9,8 cm
\item \(\widehat{SRF}\) =  29\textdegree{}
\end{itemize}

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

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

 Dans le triangle VZG rectangle en V, on sait que :
\begin{itemize}
\item VG=4,1 cm
\item \(\widehat{ZGV}\) =  21\textdegree{}
\end{itemize}

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

\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 (5.294,1.124) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{WSV}\)) = \(\dfrac{WV}{WS}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{WSV}\)) =  \(\dfrac{2{,}2}{5{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{WSV}\) =  ArcTan(\(\dfrac{2{,}2}{5{,}8}\)) \(\approx\) 21\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {7,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.806,2.742) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{ZGJ}\)) =  \(\dfrac{ZJ}{GJ}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{ZGJ}\)) = \(\dfrac{5{,}9}{7{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{ZGJ}\) =  ArcSin(\(\dfrac{5{,}9}{7{,}2}\)) \(\approx\) 55\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.672) {T};
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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.472) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.403,2.563) {64°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{KST}\)) = \(\dfrac{KT}{KS}\)
\par\addvspace{.5em}
d'où tan(64\textdegree{}) =\(\dfrac{KT}{7{,}5}\)
\par\addvspace{.5em}
On a donc KT = 7,5 \(\times\) tan(64\textdegree{}) \(\approx\) 15,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.305,0.742) {29°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{FRS}\)) = \(\dfrac{FR}{SR}\)
\par\addvspace{.5em}
  d'où    cos(29\textdegree{}) =  \(\dfrac{9{,}8}{SR}\)
\par\addvspace{.5em}
 On a donc SR =  \(\dfrac{9{,}8}{\cos(29^{\circ})}\) \(\approx\) 11,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.926) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.726) {4,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.294,1.114) {21°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{VGZ}\)) = \(\dfrac{VZ}{VG}\)
\par\addvspace{.5em}
d'où  tan(21\textdegree{}) =  \(\dfrac{VZ}{4{,}1}\)
\par\addvspace{.5em}
On a donc VZ =  4,1 \(\times\) tan(21\textdegree{})  \(\approx\) 1,6 cm\end{document}
