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\begin{document}

\sloppy

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

 Dans le triangle MCK rectangle en M, on sait que :
\begin{itemize}
\item MC=0,8 cm
\item \(\widehat{CKM}\) =  11\textdegree{}
\end{itemize}

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

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

 Dans le triangle SDN rectangle en S, on sait que :
\begin{itemize}
\item SD=2,6 cm
\item DN=7,4 cm
\end{itemize}

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

 Dans le triangle NHR rectangle en N, on sait que :
\begin{itemize}
\item NR=2,3 cm
\item \(\widehat{HRN}\) =  37\textdegree{}
\end{itemize}

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

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

 Dans le triangle TZS rectangle en T, on sait que :
\begin{itemize}
\item TZ=1,8 cm
\item TS=5,1 cm
\end{itemize}

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

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

 Dans le triangle TMS rectangle en T, on sait que :
\begin{itemize}
\item MS=2,4 cm
\item \(\widehat{TMS}\) = 58\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{MKC}\)) = \(\dfrac{MC}{MK}\)
\par\addvspace{.5em}
d'où  tan(11\textdegree{}) =  \(\dfrac{0{,}8}{MK}\)
\par\addvspace{.5em}
On a donc MK = \(\dfrac{0{,}8}{\tan(11^{\circ})}\) \(\approx\) 4,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{SDN}\)) = \(\dfrac{SD}{DN}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{SDN}\)) = \(\dfrac{2{,}6}{7{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{SDN}\) =ArcCos(\(\dfrac{2{,}6}{7{,}4}\)) \(\approx\)69\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {?};
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\end{tikzpicture}\end{center}


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

cos(\(\widehat{NRH}\)) = \(\dfrac{NR}{HR}\)
\par\addvspace{.5em}
  d'où    cos(37\textdegree{}) =  \(\dfrac{2{,}3}{HR}\)
\par\addvspace{.5em}
 On a donc HR =  \(\dfrac{2{,}3}{\cos(37^{\circ})}\) \(\approx\) 2,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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 Dans le triangle TZS rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TSZ}\) son coté opposé et son coté adjacent.

\par\addvspace{.5em}
tan(\(\widehat{TSZ}\)) = \(\dfrac{TZ}{TS}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{TSZ}\)) =  \(\dfrac{1{,}8}{5{,}1}\)
\par\addvspace{.5em}
 On a donc \(\widehat{TSZ}\) =  ArcTan(\(\dfrac{1{,}8}{5{,}1}\)) \(\approx\) 19\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.760,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {2,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.583,2.750) {58°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{TMS}\)) = \(\dfrac{TM}{MS}\)
\par\addvspace{.5em}
d'où cos(58\textdegree{}) = \(\dfrac{TM}{2{,}4}\)

On a donc TM = 2,4 \(\times\) cos(58\textdegree{}) \(\approx\) 1,3 cm\end{document}
