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\begin{document}

\sloppy

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

 Dans le triangle NTK rectangle en N, on sait que :
\begin{itemize}
\item NT=6,4 cm
\item \(\widehat{TKN}\) =  32\textdegree{}
\end{itemize}

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

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

 Dans le triangle NTH rectangle en N, on sait que :
\begin{itemize}
\item NH=5,4 cm
\item \(\widehat{NTH}\) = 72\textdegree{}
\end{itemize}

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

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

 Dans le triangle PZF rectangle en P, on sait que :
\begin{itemize}
\item ZF=1 cm
\item \(\widehat{PZF}\) = 63\textdegree{}
\end{itemize}

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

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

 Dans le triangle CBL rectangle en C, on sait que :
\begin{itemize}
\item CL=6,1 cm
\item BL=9,5 cm
\end{itemize}

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

 Dans le triangle NMR rectangle en N, on sait que :
\begin{itemize}
\item NM=1,8 cm
\item MR=7,5 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
sin(\(\widehat{NKT}\)) = \(\dfrac{NT}{TK}\)
\par\addvspace{.5em}
 D'où sin(32\textdegree{}) = \(\dfrac{6{,}4}{TK}\)
\par\addvspace{.5em}
On a donc TK =\(\dfrac{6{,}4}{\sin(32^{\circ})}\) \(\approx\) 12,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{NTH}\)) = \(\dfrac{NH}{NT}\)
\par\addvspace{.5em}
tan(72\textdegree{}) = \(\dfrac{5{,}4}{NT}\)
\par\addvspace{.5em}
 On a donc NT = \(\dfrac{5{,}4}{\tan(72^{\circ})}\) \(\approx\) 1,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{PZF}\)) =  \(\dfrac{PF}{ZF}\)
\par\addvspace{.5em}
 d'où sin(63\textdegree{}) = \(\dfrac{PF}{1}\)

On a donc PF = 1 \(\times\) sin(63\textdegree{}) \(\approx\) 0,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{CBL}\)) =  \(\dfrac{CL}{BL}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{CBL}\)) = \(\dfrac{6{,}1}{9{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{CBL}\) =  ArcSin(\(\dfrac{6{,}1}{9{,}5}\)) \(\approx\) 40\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.288,1.408) {?};
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 Dans le triangle NMR rectangle en N, on cherche une relation entre l'angle aigu   \(\widehat{NRM}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{NRM}\)) = \(\dfrac{NM}{MR}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{NRM}\)) =\(\dfrac{1{,}8}{7{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{NRM}\) = ArcSin(\(\dfrac{1{,}8}{7{,}5}\)) \(\approx\) 14\textdegree{}\end{document}
