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\begin{document}

\sloppy

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

 Dans le triangle NKW rectangle en N, on sait que :
\begin{itemize}
\item NK=2,5 cm
\item KW=9,7 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{NKW}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle PCB rectangle en P, on sait que :
\begin{itemize}
\item CB=6,7 cm
\item \(\widehat{PCB}\) = 65\textdegree{}
\end{itemize}

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

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

 Dans le triangle RBF rectangle en R, on sait que :
\begin{itemize}
\item RF=9,4 cm
\item \(\widehat{BFR}\) =  26\textdegree{}
\end{itemize}

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

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

 Dans le triangle BLZ rectangle en B, on sait que :
\begin{itemize}
\item BZ=8,3 cm
\item \(\widehat{BLZ}\) = 65\textdegree{}
\end{itemize}

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

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

 Dans le triangle CFW rectangle en C, on sait que :
\begin{itemize}
\item CF=3,1 cm
\item CW=5,4 cm
\end{itemize}

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

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

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

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.456,2.058) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{NKW}\)) = \(\dfrac{NK}{KW}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{NKW}\)) = \(\dfrac{2{,}5}{9{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{NKW}\) =ArcCos(\(\dfrac{2{,}5}{9{,}7}\)) \(\approx\)75\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{PCB}\)) =  \(\dfrac{PB}{CB}\)
\par\addvspace{.5em}
 d'où sin(65\textdegree{}) = \(\dfrac{PB}{6{,}7}\)

On a donc PB = 6,7 \(\times\) sin(65\textdegree{}) \(\approx\) 6,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{RFB}\)) = \(\dfrac{RB}{RF}\)
\par\addvspace{.5em}
d'où  tan(26\textdegree{}) =  \(\dfrac{RB}{9{,}4}\)
\par\addvspace{.5em}
On a donc RB =  9,4 \(\times\) tan(26\textdegree{})  \(\approx\) 4,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{BLZ}\)) =  \(\dfrac{BZ}{LZ}\)
\par\addvspace{.5em}
 d'où sin(65\textdegree{}) = \(\dfrac{8{,}3}{LZ}\)
\par\addvspace{.5em}
 On a donc LZ = \(\dfrac{8{,}3}{\sin(65^{\circ})}\) \(\approx\) 9,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.039,0.467) {W};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.561,1.933) {3,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {5,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.295,0.705) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{CWF}\)) = \(\dfrac{CF}{CW}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{CWF}\)) =  \(\dfrac{3{,}1}{5{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{CWF}\) =  ArcTan(\(\dfrac{3{,}1}{5{,}4}\)) \(\approx\) 30\textdegree{}\end{document}
