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\begin{document}

\sloppy

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

 Dans le triangle CNM rectangle en C, on sait que :
\begin{itemize}
\item CN=6,5 cm
\item \(\widehat{NMC}\) =  22\textdegree{}
\end{itemize}

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

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

 Dans le triangle FHK rectangle en F, on sait que :
\begin{itemize}
\item HK=8,6 cm
\item \(\widehat{HKF}\) =  14\textdegree{}
\end{itemize}

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

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

 Dans le triangle NMJ rectangle en N, on sait que :
\begin{itemize}
\item NJ=6,1 cm
\item \(\widehat{NMJ}\) = 67\textdegree{}
\end{itemize}

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

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

 Dans le triangle JZK rectangle en J, on sait que :
\begin{itemize}
\item JZ=1,2 cm
\item ZK=9,1 cm
\end{itemize}

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

 Dans le triangle CKR rectangle en C, on sait que :
\begin{itemize}
\item CK=2,1 cm
\item CR=4,2 cm
\end{itemize}

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

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

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

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{CMN}\)) = \(\dfrac{CN}{CM}\)
\par\addvspace{.5em}
d'où  tan(22\textdegree{}) =  \(\dfrac{6{,}5}{CM}\)
\par\addvspace{.5em}
On a donc CM = \(\dfrac{6{,}5}{\tan(22^{\circ})}\) \(\approx\) 16,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{FKH}\)) = \(\dfrac{FH}{HK}\)
\par\addvspace{.5em}
 D'où sin(14\textdegree{}) = \(\dfrac{FH}{8{,}6}\)

On a donc FH = 8,6 \(\times\) sin(14\textdegree{}) \(\approx\) 2,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{NMJ}\)) =  \(\dfrac{NJ}{MJ}\)
\par\addvspace{.5em}
 d'où sin(67\textdegree{}) = \(\dfrac{6{,}1}{MJ}\)
\par\addvspace{.5em}
 On a donc MJ = \(\dfrac{6{,}1}{\sin(67^{\circ})}\) \(\approx\) 6,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.489,1.758) {?};
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 Dans le triangle JZK rectangle en J, on cherche une relation entre l'angle aigu   \(\widehat{JZK}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{JZK}\)) = \(\dfrac{JZ}{ZK}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{JZK}\)) = \(\dfrac{1{,}2}{9{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{JZK}\) =ArcCos(\(\dfrac{1{,}2}{9{,}1}\)) \(\approx\)82\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \draw[black] (1.050,1.020) -- (1.130,1.020) -- (1.215,1.025) -- (1.295,1.020);
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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.642) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.442) {4,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.301,0.862) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{CRK}\)) = \(\dfrac{CK}{CR}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{CRK}\)) =  \(\dfrac{2{,}1}{4{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{CRK}\) =  ArcTan(\(\dfrac{2{,}1}{4{,}2}\)) \(\approx\) 27\textdegree{}\end{document}
