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\begin{document}

\sloppy

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

 Dans le triangle FLR rectangle en F, on sait que :
\begin{itemize}
\item FL=8,3 cm
\item \(\widehat{FLR}\) = 62\textdegree{}
\end{itemize}

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

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

 Dans le triangle KNZ rectangle en K, on sait que :
\begin{itemize}
\item KZ=5,3 cm
\item \(\widehat{KNZ}\) = 61\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZGA rectangle en Z, on sait que :
\begin{itemize}
\item ZG=1,9 cm
\item ZA=4,2 cm
\end{itemize}

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

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

 Dans le triangle DBH rectangle en D, on sait que :
\begin{itemize}
\item DB=1,4 cm
\item BH=9,6 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{DBH}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle CSV rectangle en C, on sait que :
\begin{itemize}
\item CV=0,7 cm
\item \(\widehat{CSV}\) = 49\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{FLR}\)) = \(\dfrac{FR}{FL}\)
\par\addvspace{.5em}
d'où tan(62\textdegree{}) =\(\dfrac{FR}{8{,}3}\)
\par\addvspace{.5em}
On a donc FR = 8,3 \(\times\) tan(62\textdegree{}) \(\approx\) 15,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{KNZ}\)) =  \(\dfrac{KZ}{NZ}\)
\par\addvspace{.5em}
 d'où sin(61\textdegree{}) = \(\dfrac{5{,}3}{NZ}\)
\par\addvspace{.5em}
 On a donc NZ = \(\dfrac{5{,}3}{\sin(61^{\circ})}\) \(\approx\) 6,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{ZAG}\)) = \(\dfrac{ZG}{ZA}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{ZAG}\)) =  \(\dfrac{1{,}9}{4{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ZAG}\) =  ArcTan(\(\dfrac{1{,}9}{4{,}2}\)) \(\approx\) 24\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{DBH}\)) = \(\dfrac{DB}{BH}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{DBH}\)) = \(\dfrac{1{,}4}{9{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{DBH}\) =ArcCos(\(\dfrac{1{,}4}{9{,}6}\)) \(\approx\)82\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {0,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.166,2.727) {49°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{CSV}\)) = \(\dfrac{CV}{CS}\)
\par\addvspace{.5em}
tan(49\textdegree{}) = \(\dfrac{0{,}7}{CS}\)
\par\addvspace{.5em}
 On a donc CS = \(\dfrac{0{,}7}{\tan(49^{\circ})}\) \(\approx\) 0,6 cm\end{document}
