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\begin{document}

\sloppy

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

 Dans le triangle FHV rectangle en F, on sait que :
\begin{itemize}
\item FH=2,3 cm
\item HV=9,7 cm
\end{itemize}

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

 Dans le triangle FZD rectangle en F, on sait que :
\begin{itemize}
\item ZD=0,6 cm
\item \(\widehat{FZD}\) = 48\textdegree{}
\end{itemize}

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

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

 Dans le triangle GDK rectangle en G, on sait que :
\begin{itemize}
\item GD=1 cm
\item GK=6,5 cm
\end{itemize}

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

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

 Dans le triangle KSH rectangle en K, on sait que :
\begin{itemize}
\item KH=6,3 cm
\item \(\widehat{KSH}\) = 52\textdegree{}
\end{itemize}

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

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

 Dans le triangle KBZ rectangle en K, on sait que :
\begin{itemize}
\item KZ=5,2 cm
\item \(\widehat{KBZ}\) = 57\textdegree{}
\end{itemize}

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

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

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

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{FHV}\)) = \(\dfrac{FH}{HV}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{FHV}\)) = \(\dfrac{2{,}3}{9{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{FHV}\) =ArcCos(\(\dfrac{2{,}3}{9{,}7}\)) \(\approx\)76\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{FZD}\)) =  \(\dfrac{FD}{ZD}\)
\par\addvspace{.5em}
 d'où sin(48\textdegree{}) = \(\dfrac{FD}{0{,}6}\)

On a donc FD = 0,6 \(\times\) sin(48\textdegree{}) \(\approx\) 0,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{GKD}\)) = \(\dfrac{GD}{GK}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{GKD}\)) =  \(\dfrac{1}{6{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{GKD}\) =  ArcTan(\(\dfrac{1}{6{,}5}\)) \(\approx\) 9\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{KSH}\)) = \(\dfrac{KH}{KS}\)
\par\addvspace{.5em}
tan(52\textdegree{}) = \(\dfrac{6{,}3}{KS}\)
\par\addvspace{.5em}
 On a donc KS = \(\dfrac{6{,}3}{\tan(52^{\circ})}\) \(\approx\) 4,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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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 (1.662,2.747) {57°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{KBZ}\)) =  \(\dfrac{KZ}{BZ}\)
\par\addvspace{.5em}
 d'où sin(57\textdegree{}) = \(\dfrac{5{,}2}{BZ}\)
\par\addvspace{.5em}
 On a donc BZ = \(\dfrac{5{,}2}{\sin(57^{\circ})}\) \(\approx\) 6,2 cm\end{document}
