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\begin{document}

\sloppy

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

 Dans le triangle HTC rectangle en H, on sait que :
\begin{itemize}
\item HC=1,2 cm
\item \(\widehat{TCH}\) =  29\textdegree{}
\end{itemize}

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

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

 Dans le triangle PZT rectangle en P, on sait que :
\begin{itemize}
\item PZ=6,3 cm
\item \(\widehat{PZT}\) = 62\textdegree{}
\end{itemize}

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

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

 Dans le triangle HTK rectangle en H, on sait que :
\begin{itemize}
\item HT=1,6 cm
\item TK=7,4 cm
\end{itemize}

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

 Dans le triangle MDG rectangle en M, on sait que :
\begin{itemize}
\item DG=6,3 cm
\item \(\widehat{DGM}\) =  45\textdegree{}
\end{itemize}

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

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

 Dans le triangle FJN rectangle en F, on sait que :
\begin{itemize}
\item FJ=3,2 cm
\item FN=5,9 cm
\end{itemize}

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

\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 (5.305,0.742) {29°};
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 Dans le triangle HTC rectangle en H, on cherche une relation entre l'angle aigu   \(\widehat{HCT}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{HCT}\)) = \(\dfrac{HC}{TC}\)
\par\addvspace{.5em}
  d'où    cos(29\textdegree{}) =  \(\dfrac{1{,}2}{TC}\)
\par\addvspace{.5em}
 On a donc TC =  \(\dfrac{1{,}2}{\cos(29^{\circ})}\) \(\approx\) 1,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{PZT}\)) = \(\dfrac{PT}{PZ}\)
\par\addvspace{.5em}
d'où tan(62\textdegree{}) =\(\dfrac{PT}{6{,}3}\)
\par\addvspace{.5em}
On a donc PT = 6,3 \(\times\) tan(62\textdegree{}) \(\approx\) 11,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{HTK}\)) = \(\dfrac{HT}{TK}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{HTK}\)) = \(\dfrac{1{,}6}{7{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{HTK}\) =ArcCos(\(\dfrac{1{,}6}{7{,}4}\)) \(\approx\)78\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{MGD}\)) = \(\dfrac{MD}{DG}\)
\par\addvspace{.5em}
 D'où sin(45\textdegree{}) = \(\dfrac{MD}{6{,}3}\)

On a donc MD = 6,3 \(\times\) sin(45\textdegree{}) \(\approx\) 4,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.338) {5,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.304,0.768) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{FNJ}\)) = \(\dfrac{FJ}{FN}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{FNJ}\)) =  \(\dfrac{3{,}2}{5{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{FNJ}\) =  ArcTan(\(\dfrac{3{,}2}{5{,}9}\)) \(\approx\) 28\textdegree{}\end{document}
