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\begin{document}

\sloppy

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

 Dans le triangle HNA rectangle en H, on sait que :
\begin{itemize}
\item HN=0,9 cm
\item \(\widehat{HNA}\) = 72\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZTH rectangle en Z, on sait que :
\begin{itemize}
\item ZH=6,4 cm
\item TH=7,6 cm
\end{itemize}

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

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

 Dans le triangle HSL rectangle en H, on sait que :
\begin{itemize}
\item HS=2,5 cm
\item \(\widehat{SLH}\) =  23\textdegree{}
\end{itemize}

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

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

 Dans le triangle DMV rectangle en D, on sait que :
\begin{itemize}
\item DM=2,4 cm
\item DV=4,5 cm
\end{itemize}

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

 Dans le triangle KPJ rectangle en K, on sait que :
\begin{itemize}
\item KJ=5,3 cm
\item \(\widehat{PJK}\) =  20\textdegree{}
\end{itemize}

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

\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 (1.442,2.190) {72°};
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 Dans le triangle HNA rectangle en H, on cherche une relation entre l'angle aigu   \(\widehat{HNA}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{HNA}\)) = \(\dfrac{HN}{NA}\)
\par\addvspace{.5em}
 d'où cos(72\textdegree{}) = \(\dfrac{0{,}9}{NA}\)
\par\addvspace{.5em}
On a donc NA = \(\dfrac{0{,}9}{\cos(72^{\circ})}\)\(\approx\) 2,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{ZHT}\)) = \(\dfrac{ZH}{TH}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{ZHT}\)) =  \(\dfrac{6{,}4}{7{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{ZHT}\) = Arccos (\(\dfrac{6{,}4}{7{,}6}\)) \(\approx\) 33\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{HLS}\)) = \(\dfrac{HS}{HL}\)
\par\addvspace{.5em}
d'où  tan(23\textdegree{}) =  \(\dfrac{2{,}5}{HL}\)
\par\addvspace{.5em}
On a donc HL = \(\dfrac{2{,}5}{\tan(23^{\circ})}\) \(\approx\) 5,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.393,2.668) {?};
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 Dans le triangle DMV rectangle en D, on cherche une relation entre l'angle aigu   \(\widehat{DMV}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{DMV}\)) = \(\dfrac{DV}{DM}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{DMV}\)) = \(\dfrac{4{,}5}{2{,}4}\)
\par\addvspace{.5em}
 On a donc DMV = ArcTan(\(\dfrac{4{,}5}{2{,}4}\)) \(\approx\) 62\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.975) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.775) {5,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.293,1.157) {20°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{KJP}\)) = \(\dfrac{KP}{KJ}\)
\par\addvspace{.5em}
d'où  tan(20\textdegree{}) =  \(\dfrac{KP}{5{,}3}\)
\par\addvspace{.5em}
On a donc KP =  5,3 \(\times\) tan(20\textdegree{})  \(\approx\) 1,9 cm\end{document}
