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\begin{document}

\sloppy

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

 Dans le triangle CHR rectangle en C, on sait que :
\begin{itemize}
\item CR=1,2 cm
\item \(\widehat{CHR}\) = 64\textdegree{}
\end{itemize}

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

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

 Dans le triangle GNT rectangle en G, on sait que :
\begin{itemize}
\item GT=3,8 cm
\item NT=9,8 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{GNT}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle TGB rectangle en T, on sait que :
\begin{itemize}
\item GB=7 cm
\item \(\widehat{GBT}\) =  10\textdegree{}
\end{itemize}

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

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

 Dans le triangle DLS rectangle en D, on sait que :
\begin{itemize}
\item LS=4,5 cm
\item \(\widehat{LSD}\) =  31\textdegree{}
\end{itemize}

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

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

 Dans le triangle MNK rectangle en M, on sait que :
\begin{itemize}
\item MN=1,1 cm
\item MK=5,3 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{CHR}\)) =  \(\dfrac{CR}{HR}\)
\par\addvspace{.5em}
 d'où sin(64\textdegree{}) = \(\dfrac{1{,}2}{HR}\)
\par\addvspace{.5em}
 On a donc HR = \(\dfrac{1{,}2}{\sin(64^{\circ})}\) \(\approx\) 1,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{GNT}\)) =  \(\dfrac{GT}{NT}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{GNT}\)) = \(\dfrac{3{,}8}{9{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{GNT}\) =  ArcSin(\(\dfrac{3{,}8}{9{,}8}\)) \(\approx\) 23\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{TBG}\)) = \(\dfrac{TG}{GB}\)
\par\addvspace{.5em}
 D'où sin(10\textdegree{}) = \(\dfrac{TG}{7}\)

On a donc TG = 7 \(\times\) sin(10\textdegree{}) \(\approx\) 1,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{DSL}\)) = \(\dfrac{DS}{LS}\)
\par\addvspace{.5em}
  d'où    cos(31\textdegree{}) =  \(\dfrac{DS}{4{,}5}\)

On a donc DS = 4,5 \(\times\) cos(31\textdegree{}) \(\approx\) 3{,}9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.287,1.493) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{MKN}\)) = \(\dfrac{MN}{MK}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{MKN}\)) =  \(\dfrac{1{,}1}{5{,}3}\)
\par\addvspace{.5em}
 On a donc \(\widehat{MKN}\) =  ArcTan(\(\dfrac{1{,}1}{5{,}3}\)) \(\approx\) 12\textdegree{}\end{document}
