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\begin{document}

\sloppy

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

 Dans le triangle MDL rectangle en M, on sait que :
\begin{itemize}
\item MD=6,2 cm
\item \(\widehat{MDL}\) = 69\textdegree{}
\end{itemize}

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

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

 Dans le triangle TBD rectangle en T, on sait que :
\begin{itemize}
\item TB=5,5 cm
\item \(\widehat{TBD}\) = 64\textdegree{}
\end{itemize}

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

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

 Dans le triangle NJC rectangle en N, on sait que :
\begin{itemize}
\item NC=8,2 cm
\item \(\widehat{NJC}\) = 56\textdegree{}
\end{itemize}

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

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

 Dans le triangle GHF rectangle en G, on sait que :
\begin{itemize}
\item GH=1 cm
\item HF=9,9 cm
\end{itemize}

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

 Dans le triangle TZC rectangle en T, on sait que :
\begin{itemize}
\item TC=6,3 cm
\item ZC=9,5 cm
\end{itemize}

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

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

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

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

cos(\(\widehat{MDL}\)) = \(\dfrac{MD}{DL}\)
\par\addvspace{.5em}
 d'où cos(69\textdegree{}) = \(\dfrac{6{,}2}{DL}\)
\par\addvspace{.5em}
On a donc DL = \(\dfrac{6{,}2}{\cos(69^{\circ})}\)\(\approx\) 17,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{TBD}\)) = \(\dfrac{TD}{TB}\)
\par\addvspace{.5em}
d'où tan(64\textdegree{}) =\(\dfrac{TD}{5{,}5}\)
\par\addvspace{.5em}
On a donc TD = 5,5 \(\times\) tan(64\textdegree{}) \(\approx\) 11,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
tan(\(\widehat{NJC}\)) = \(\dfrac{NC}{NJ}\)
\par\addvspace{.5em}
tan(56\textdegree{}) = \(\dfrac{8{,}2}{NJ}\)
\par\addvspace{.5em}
 On a donc NJ = \(\dfrac{8{,}2}{\tan(56^{\circ})}\) \(\approx\) 5,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{GHF}\)) = \(\dfrac{GH}{HF}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{GHF}\)) = \(\dfrac{1}{9{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{GHF}\) =ArcCos(\(\dfrac{1}{9{,}9}\)) \(\approx\)84\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.132,0.807) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{TCZ}\)) = \(\dfrac{TC}{ZC}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{TCZ}\)) =  \(\dfrac{6{,}3}{9{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{TCZ}\) = Arccos (\(\dfrac{6{,}3}{9{,}5}\)) \(\approx\) 48\textdegree{} \end{document}
