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\begin{document}

\sloppy

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

 Dans le triangle FPD rectangle en F, on sait que :
\begin{itemize}
\item FP=6 cm
\item \(\widehat{PDF}\) =  39\textdegree{}
\end{itemize}

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

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

 Dans le triangle TRL rectangle en T, on sait que :
\begin{itemize}
\item TR=7,4 cm
\item \(\widehat{TRL}\) = 70\textdegree{}
\end{itemize}

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

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

 Dans le triangle TWF rectangle en T, on sait que :
\begin{itemize}
\item TF=4,6 cm
\item \(\widehat{WFT}\) =  26\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZRG rectangle en Z, on sait que :
\begin{itemize}
\item ZR=1,6 cm
\item RG=8,1 cm
\end{itemize}

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

 Dans le triangle WZM rectangle en W, on sait que :
\begin{itemize}
\item WZ=3,3 cm
\item WM=4,7 cm
\end{itemize}

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

\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 (4.600,0.756) {39°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{FDP}\)) = \(\dfrac{FP}{FD}\)
\par\addvspace{.5em}
d'où  tan(39\textdegree{}) =  \(\dfrac{6}{FD}\)
\par\addvspace{.5em}
On a donc FD = \(\dfrac{6}{\tan(39^{\circ})}\) \(\approx\) 7,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{TRL}\)) = \(\dfrac{TR}{RL}\)
\par\addvspace{.5em}
 d'où cos(70\textdegree{}) = \(\dfrac{7{,}4}{RL}\)
\par\addvspace{.5em}
On a donc RL = \(\dfrac{7{,}4}{\cos(70^{\circ})}\)\(\approx\) 21,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{TFW}\)) = \(\dfrac{TW}{TF}\)
\par\addvspace{.5em}
d'où  tan(26\textdegree{}) =  \(\dfrac{TW}{4{,}6}\)
\par\addvspace{.5em}
On a donc TW =  4,6 \(\times\) tan(26\textdegree{})  \(\approx\) 2,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{ZRG}\)) = \(\dfrac{ZR}{RG}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{ZRG}\)) = \(\dfrac{1{,}6}{8{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{ZRG}\) =ArcCos(\(\dfrac{1{,}6}{8{,}1}\)) \(\approx\)79\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.594,0.467) {M};
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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {4,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.858,0.734) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{WMZ}\)) = \(\dfrac{WZ}{WM}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{WMZ}\)) =  \(\dfrac{3{,}3}{4{,}7}\)
\par\addvspace{.5em}
 On a donc \(\widehat{WMZ}\) =  ArcTan(\(\dfrac{3{,}3}{4{,}7}\)) \(\approx\) 35\textdegree{}\end{document}
