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\begin{document}

\sloppy

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

 Dans le triangle HMD rectangle en H, on sait que :
\begin{itemize}
\item HD=7,4 cm
\item \(\widehat{HMD}\) = 78\textdegree{}
\end{itemize}

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

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

 Dans le triangle HTZ rectangle en H, on sait que :
\begin{itemize}
\item HZ=4,7 cm
\item TZ=8 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{HTZ}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle ARK rectangle en A, on sait que :
\begin{itemize}
\item AR=5,3 cm
\item \(\widehat{ARK}\) = 60\textdegree{}
\end{itemize}

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

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

 Dans le triangle LJD rectangle en L, on sait que :
\begin{itemize}
\item LD=7,6 cm
\item \(\widehat{LJD}\) = 67\textdegree{}
\end{itemize}

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

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

 Dans le triangle KAC rectangle en K, on sait que :
\begin{itemize}
\item KA=1,3 cm
\item KC=4,1 cm
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{HMD}\)) = \(\dfrac{HD}{HM}\)
\par\addvspace{.5em}
tan(78\textdegree{}) = \(\dfrac{7{,}4}{HM}\)
\par\addvspace{.5em}
 On a donc HM = \(\dfrac{7{,}4}{\tan(78^{\circ})}\) \(\approx\) 1,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
sin(\(\widehat{HTZ}\)) =  \(\dfrac{HZ}{TZ}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{HTZ}\)) = \(\dfrac{4{,}7}{8}\)
\par\addvspace{.5em}
On a donc \(\widehat{HTZ}\) =  ArcSin(\(\dfrac{4{,}7}{8}\)) \(\approx\) 36\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{ARK}\)) = \(\dfrac{AK}{AR}\)
\par\addvspace{.5em}
d'où tan(60\textdegree{}) =\(\dfrac{AK}{5{,}3}\)
\par\addvspace{.5em}
On a donc AK = 5,3 \(\times\) tan(60\textdegree{}) \(\approx\) 9,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{LJD}\)) =  \(\dfrac{LD}{JD}\)
\par\addvspace{.5em}
 d'où sin(67\textdegree{}) = \(\dfrac{7{,}6}{JD}\)
\par\addvspace{.5em}
 On a donc JD = \(\dfrac{7{,}6}{\sin(67^{\circ})}\) \(\approx\) 8,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.890) {4,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.291,1.258) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{KCA}\)) = \(\dfrac{KA}{KC}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{KCA}\)) =  \(\dfrac{1{,}3}{4{,}1}\)
\par\addvspace{.5em}
 On a donc \(\widehat{KCA}\) =  ArcTan(\(\dfrac{1{,}3}{4{,}1}\)) \(\approx\) 18\textdegree{}\end{document}
