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\begin{document}

\sloppy

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

 Dans le triangle KSD rectangle en K, on sait que :
\begin{itemize}
\item KS=8,3 cm
\item \(\widehat{KSD}\) = 48\textdegree{}
\end{itemize}

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

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

 Dans le triangle RGM rectangle en R, on sait que :
\begin{itemize}
\item RG=2,4 cm
\item RM=3,5 cm
\end{itemize}

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

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

 Dans le triangle KZG rectangle en K, on sait que :
\begin{itemize}
\item KZ=2,1 cm
\item ZG=9 cm
\end{itemize}

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

 Dans le triangle VTH rectangle en V, on sait que :
\begin{itemize}
\item TH=3,6 cm
\item \(\widehat{VTH}\) = 79\textdegree{}
\end{itemize}

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

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

 Dans le triangle CBA rectangle en C, on sait que :
\begin{itemize}
\item CB=6,5 cm
\item \(\widehat{BAC}\) =  14\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [AB]. (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 (2.216,2.724) {48°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{KSD}\)) = \(\dfrac{KD}{KS}\)
\par\addvspace{.5em}
d'où tan(48\textdegree{}) =\(\dfrac{KD}{8{,}3}\)
\par\addvspace{.5em}
On a donc KD = 8,3 \(\times\) tan(48\textdegree{}) \(\approx\) 9,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{RMG}\)) = \(\dfrac{RG}{RM}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{RMG}\)) =  \(\dfrac{2{,}4}{3{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{RMG}\) =  ArcTan(\(\dfrac{2{,}4}{3{,}5}\)) \(\approx\) 34\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{KZG}\)) = \(\dfrac{KZ}{ZG}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{KZG}\)) = \(\dfrac{2{,}1}{9}\)
\par\addvspace{.5em}
On a donc \(\widehat{KZG}\) =ArcCos(\(\dfrac{2{,}1}{9}\)) \(\approx\)77\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.474,1.895) {79°};
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 Dans le triangle VTH rectangle en V, on cherche une relation entre l'angle aigu   \(\widehat{VTH}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{VTH}\)) = \(\dfrac{VT}{TH}\)
\par\addvspace{.5em}
d'où cos(79\textdegree{}) = \(\dfrac{VT}{3{,}6}\)

On a donc VT = 3,6 \(\times\) cos(79\textdegree{}) \(\approx\) 0,7 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 (5.288,1.404) {14°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{CAB}\)) = \(\dfrac{CB}{BA}\)
\par\addvspace{.5em}
 D'où sin(14\textdegree{}) = \(\dfrac{6{,}5}{BA}\)
\par\addvspace{.5em}
On a donc BA =\(\dfrac{6{,}5}{\sin(14^{\circ})}\) \(\approx\) 26,9 cm\end{document}
