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\begin{document}

\sloppy

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

 Dans le triangle TZH rectangle en T, on sait que :
\begin{itemize}
\item TZ=1 cm
\item TH=4,3 cm
\end{itemize}

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

 Dans le triangle RGF rectangle en R, on sait que :
\begin{itemize}
\item RF=9,2 cm
\item \(\widehat{RGF}\) = 61\textdegree{}
\end{itemize}

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

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

 Dans le triangle DHZ rectangle en D, on sait que :
\begin{itemize}
\item HZ=0,5 cm
\item \(\widehat{HZD}\) =  38\textdegree{}
\end{itemize}

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

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

 Dans le triangle WLV rectangle en W, on sait que :
\begin{itemize}
\item WL=1,6 cm
\item WV=5,8 cm
\end{itemize}

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

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

 Dans le triangle WRF rectangle en W, on sait que :
\begin{itemize}
\item WR=6,1 cm
\item \(\widehat{RFW}\) =  40\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{TZH}\)) = \(\dfrac{TH}{TZ}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{TZH}\)) = \(\dfrac{4{,}3}{1}\)
\par\addvspace{.5em}
 On a donc TZH = ArcTan(4,3) \(\approx\) 77\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{RGF}\)) = \(\dfrac{RF}{RG}\)
\par\addvspace{.5em}
tan(61\textdegree{}) = \(\dfrac{9{,}2}{RG}\)
\par\addvspace{.5em}
 On a donc RG = \(\dfrac{9{,}2}{\tan(61^{\circ})}\) \(\approx\) 5,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{DZH}\)) = \(\dfrac{DZ}{HZ}\)
\par\addvspace{.5em}
  d'où    cos(38\textdegree{}) =  \(\dfrac{DZ}{0{,}5}\)

On a donc DZ = 0,5 \(\times\) cos(38\textdegree{}) \(\approx\) 0{,}4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{WVL}\)) = \(\dfrac{WL}{WV}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{WVL}\)) =  \(\dfrac{1{,}6}{5{,}8}\)
\par\addvspace{.5em}
 On a donc \(\widehat{WVL}\) =  ArcTan(\(\dfrac{1{,}6}{5{,}8}\)) \(\approx\) 15\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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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 (4.542,0.761) {40°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{WFR}\)) = \(\dfrac{WR}{RF}\)
\par\addvspace{.5em}
 D'où sin(40\textdegree{}) = \(\dfrac{6{,}1}{RF}\)
\par\addvspace{.5em}
On a donc RF =\(\dfrac{6{,}1}{\sin(40^{\circ})}\) \(\approx\) 9,5 cm\end{document}
