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\begin{document}

\sloppy

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

 Dans le triangle CKF rectangle en C, on sait que :
\begin{itemize}
\item CK=2,6 cm
\item KF=9,5 cm
\end{itemize}

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

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

 Dans le triangle THM rectangle en T, on sait que :
\begin{itemize}
\item TH=1,6 cm
\item HM=9,5 cm
\end{itemize}

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

 Dans le triangle BJR rectangle en B, on sait que :
\begin{itemize}
\item BJ=6,8 cm
\item \(\widehat{JRB}\) =  21\textdegree{}
\end{itemize}

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

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

 Dans le triangle CML rectangle en C, on sait que :
\begin{itemize}
\item CM=9,8 cm
\item \(\widehat{MLC}\) =  25\textdegree{}
\end{itemize}

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

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

 Dans le triangle BCN rectangle en B, on sait que :
\begin{itemize}
\item CN=7,5 cm
\item \(\widehat{CNB}\) =  15\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{CFK}\)) = \(\dfrac{CK}{KF}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{CFK}\)) =\(\dfrac{2{,}6}{9{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{CFK}\) = ArcSin(\(\dfrac{2{,}6}{9{,}5}\)) \(\approx\) 16\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{THM}\)) = \(\dfrac{TH}{HM}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{THM}\)) = \(\dfrac{1{,}6}{9{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{THM}\) =ArcCos(\(\dfrac{1{,}6}{9{,}5}\)) \(\approx\)80\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) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.294,1.114) {21°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{BRJ}\)) = \(\dfrac{BJ}{JR}\)
\par\addvspace{.5em}
 D'où sin(21\textdegree{}) = \(\dfrac{6{,}8}{JR}\)
\par\addvspace{.5em}
On a donc JR =\(\dfrac{6{,}8}{\sin(21^{\circ})}\) \(\approx\) 19 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{CLM}\)) = \(\dfrac{CM}{CL}\)
\par\addvspace{.5em}
d'où  tan(25\textdegree{}) =  \(\dfrac{9{,}8}{CL}\)
\par\addvspace{.5em}
On a donc CL = \(\dfrac{9{,}8}{\tan(25^{\circ})}\) \(\approx\) 21 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.210) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {7,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.289,1.364) {15°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{BNC}\)) = \(\dfrac{BC}{CN}\)
\par\addvspace{.5em}
 D'où sin(15\textdegree{}) = \(\dfrac{BC}{7{,}5}\)

On a donc BC = 7,5 \(\times\) sin(15\textdegree{}) \(\approx\) 1,9 cm\end{document}
