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\begin{document}

\sloppy

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

 Dans le triangle TCH rectangle en T, on sait que :
\begin{itemize}
\item CH=7,3 cm
\item \(\widehat{TCH}\) = 71\textdegree{}
\end{itemize}

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

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

 Dans le triangle CKV rectangle en C, on sait que :
\begin{itemize}
\item CV=4,6 cm
\item \(\widehat{CKV}\) = 60\textdegree{}
\end{itemize}

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

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

 Dans le triangle VKC rectangle en V, on sait que :
\begin{itemize}
\item VC=4,9 cm
\item \(\widehat{VKC}\) = 51\textdegree{}
\end{itemize}

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

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

 Dans le triangle RDV rectangle en R, on sait que :
\begin{itemize}
\item RD=3,3 cm
\item RV=4,1 cm
\end{itemize}

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

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

 Dans le triangle WZR rectangle en W, on sait que :
\begin{itemize}
\item WZ=3 cm
\item ZR=8,4 cm
\end{itemize}

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{TCH}\)) =  \(\dfrac{TH}{CH}\)
\par\addvspace{.5em}
 d'où sin(71\textdegree{}) = \(\dfrac{TH}{7{,}3}\)

On a donc TH = 7,3 \(\times\) sin(71\textdegree{}) \(\approx\) 6,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{CKV}\)) = \(\dfrac{CV}{CK}\)
\par\addvspace{.5em}
tan(60\textdegree{}) = \(\dfrac{4{,}6}{CK}\)
\par\addvspace{.5em}
 On a donc CK = \(\dfrac{4{,}6}{\tan(60^{\circ})}\) \(\approx\) 2,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{VKC}\)) =  \(\dfrac{VC}{KC}\)
\par\addvspace{.5em}
 d'où sin(51\textdegree{}) = \(\dfrac{4{,}9}{KC}\)
\par\addvspace{.5em}
 On a donc KC = \(\dfrac{4{,}9}{\sin(51^{\circ})}\) \(\approx\) 6,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{RVD}\)) = \(\dfrac{RD}{RV}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{RVD}\)) =  \(\dfrac{3{,}3}{4{,}1}\)
\par\addvspace{.5em}
 On a donc \(\widehat{RVD}\) =  ArcTan(\(\dfrac{3{,}3}{4{,}1}\)) \(\approx\) 39\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.428,2.321) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{WZR}\)) = \(\dfrac{WZ}{ZR}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{WZR}\)) = \(\dfrac{3}{8{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{WZR}\) =ArcCos(\(\dfrac{3}{8{,}4}\)) \(\approx\)69\textdegree{}\end{document}
