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\begin{document}

\sloppy

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

 Dans le triangle TZN rectangle en T, on sait que :
\begin{itemize}
\item TZ=3,2 cm
\item ZN=7,6 cm
\end{itemize}

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

 Dans le triangle TRK rectangle en T, on sait que :
\begin{itemize}
\item TR=1,4 cm
\item \(\widehat{RKT}\) =  40\textdegree{}
\end{itemize}

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

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

 Dans le triangle DAG rectangle en D, on sait que :
\begin{itemize}
\item DG=6,2 cm
\item \(\widehat{DAG}\) = 68\textdegree{}
\end{itemize}

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

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

 Dans le triangle HFJ rectangle en H, on sait que :
\begin{itemize}
\item HF=1,6 cm
\item HJ=4 cm
\end{itemize}

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

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

 Dans le triangle FGW rectangle en F, on sait que :
\begin{itemize}
\item FW=5,7 cm
\item \(\widehat{GWF}\) =  19\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{TZN}\)) = \(\dfrac{TZ}{ZN}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{TZN}\)) = \(\dfrac{3{,}2}{7{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{TZN}\) =ArcCos(\(\dfrac{3{,}2}{7{,}6}\)) \(\approx\)65\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{TKR}\)) = \(\dfrac{TR}{TK}\)
\par\addvspace{.5em}
d'où  tan(40\textdegree{}) =  \(\dfrac{1{,}4}{TK}\)
\par\addvspace{.5em}
On a donc TK = \(\dfrac{1{,}4}{\tan(40^{\circ})}\) \(\approx\) 1,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{DAG}\)) = \(\dfrac{DG}{DA}\)
\par\addvspace{.5em}
tan(68\textdegree{}) = \(\dfrac{6{,}2}{DA}\)
\par\addvspace{.5em}
 On a donc DA = \(\dfrac{6{,}2}{\tan(68^{\circ})}\) \(\approx\) 2,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{HJF}\)) = \(\dfrac{HF}{HJ}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{HJF}\)) =  \(\dfrac{1{,}6}{4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{HJF}\) =  ArcTan(\(\dfrac{1{,}6}{4}\)) \(\approx\) 22\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 (5.292,1.200) {19°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{FWG}\)) = \(\dfrac{FW}{GW}\)
\par\addvspace{.5em}
  d'où    cos(19\textdegree{}) =  \(\dfrac{5{,}7}{GW}\)
\par\addvspace{.5em}
 On a donc GW =  \(\dfrac{5{,}7}{\cos(19^{\circ})}\) \(\approx\) 6 cm\end{document}
