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\begin{document}

\sloppy

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

 Dans le triangle FNZ rectangle en F, on sait que :
\begin{itemize}
\item FZ=6 cm
\item NZ=8,1 cm
\end{itemize}

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

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

 Dans le triangle CFP rectangle en C, on sait que :
\begin{itemize}
\item FP=3,4 cm
\item \(\widehat{FPC}\) =  45\textdegree{}
\end{itemize}

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

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

 Dans le triangle TDW rectangle en T, on sait que :
\begin{itemize}
\item TW=5,9 cm
\item DW=8,3 cm
\end{itemize}

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

 Dans le triangle MCK rectangle en M, on sait que :
\begin{itemize}
\item MK=8,8 cm
\item \(\widehat{CKM}\) =  27\textdegree{}
\end{itemize}

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

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

 Dans le triangle VJG rectangle en V, on sait que :
\begin{itemize}
\item VG=1,4 cm
\item \(\widehat{JGV}\) =  15\textdegree{}
\end{itemize}

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

\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 (4.422,0.773) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{FZN}\)) = \(\dfrac{FZ}{NZ}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{FZN}\)) =  \(\dfrac{6}{8{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{FZN}\) = Arccos (\(\dfrac{6}{8{,}1}\)) \(\approx\) 42\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{CPF}\)) = \(\dfrac{CP}{FP}\)
\par\addvspace{.5em}
  d'où    cos(45\textdegree{}) =  \(\dfrac{CP}{3{,}4}\)

On a donc CP = 3,4 \(\times\) cos(45\textdegree{}) \(\approx\) 2{,}4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{TDW}\)) =  \(\dfrac{TW}{DW}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{TDW}\)) = \(\dfrac{5{,}9}{8{,}3}\)
\par\addvspace{.5em}
On a donc \(\widehat{TDW}\) =  ArcSin(\(\dfrac{5{,}9}{8{,}3}\)) \(\approx\) 45\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{MKC}\)) = \(\dfrac{MK}{CK}\)
\par\addvspace{.5em}
  d'où    cos(27\textdegree{}) =  \(\dfrac{8{,}8}{CK}\)
\par\addvspace{.5em}
 On a donc CK =  \(\dfrac{8{,}8}{\cos(27^{\circ})}\) \(\approx\) 9,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.289,1.364) {15°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{VGJ}\)) = \(\dfrac{VJ}{VG}\)
\par\addvspace{.5em}
d'où  tan(15\textdegree{}) =  \(\dfrac{VJ}{1{,}4}\)
\par\addvspace{.5em}
On a donc VJ =  1,4 \(\times\) tan(15\textdegree{})  \(\approx\) 0,4 cm\end{document}
