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\begin{document}

\sloppy

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

 Dans le triangle KCV rectangle en K, on sait que :
\begin{itemize}
\item KC=2,7 cm
\item KV=5,8 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{KCV}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle HKC rectangle en H, on sait que :
\begin{itemize}
\item HC=3,5 cm
\item \(\widehat{HKC}\) = 77\textdegree{}
\end{itemize}

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

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

 Dans le triangle FLD rectangle en F, on sait que :
\begin{itemize}
\item FL=3,2 cm
\item FD=5,9 cm
\end{itemize}

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

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

 Dans le triangle CWR rectangle en C, on sait que :
\begin{itemize}
\item CW=6,8 cm
\item \(\widehat{CWR}\) = 68\textdegree{}
\end{itemize}

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

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

 Dans le triangle JNG rectangle en J, on sait que :
\begin{itemize}
\item JN=5,7 cm
\item \(\widehat{NGJ}\) =  31\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{KCV}\)) = \(\dfrac{KV}{KC}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{KCV}\)) = \(\dfrac{5{,}8}{2{,}7}\)
\par\addvspace{.5em}
 On a donc KCV = ArcTan(\(\dfrac{5{,}8}{2{,}7}\)) \(\approx\) 65\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{HKC}\)) = \(\dfrac{HC}{HK}\)
\par\addvspace{.5em}
tan(77\textdegree{}) = \(\dfrac{3{,}5}{HK}\)
\par\addvspace{.5em}
 On a donc HK = \(\dfrac{3{,}5}{\tan(77^{\circ})}\) \(\approx\) 0,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.304,0.768) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{FDL}\)) = \(\dfrac{FL}{FD}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{FDL}\)) =  \(\dfrac{3{,}2}{5{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{FDL}\) =  ArcTan(\(\dfrac{3{,}2}{5{,}9}\)) \(\approx\) 28\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.423,2.371) {68°};
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 Dans le triangle CWR rectangle en C, on cherche une relation entre l'angle aigu   \(\widehat{CWR}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{CWR}\)) = \(\dfrac{CW}{WR}\)
\par\addvspace{.5em}
 d'où cos(68\textdegree{}) = \(\dfrac{6{,}8}{WR}\)
\par\addvspace{.5em}
On a donc WR = \(\dfrac{6{,}8}{\cos(68^{\circ})}\)\(\approx\) 18,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.930,0.467) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.670,1.933) {5,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.188,0.711) {31°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{JGN}\)) = \(\dfrac{JN}{JG}\)
\par\addvspace{.5em}
d'où  tan(31\textdegree{}) =  \(\dfrac{5{,}7}{JG}\)
\par\addvspace{.5em}
On a donc JG = \(\dfrac{5{,}7}{\tan(31^{\circ})}\) \(\approx\) 9,5 cm\end{document}
