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\begin{document}

\sloppy

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

 Dans le triangle JHZ rectangle en J, on sait que :
\begin{itemize}
\item JH=1,4 cm
\item JZ=4,7 cm
\end{itemize}

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

 Dans le triangle BNR rectangle en B, on sait que :
\begin{itemize}
\item NR=6,1 cm
\item \(\widehat{BNR}\) = 67\textdegree{}
\end{itemize}

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

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

 Dans le triangle AJD rectangle en A, on sait que :
\begin{itemize}
\item AJ=1,2 cm
\item AD=5,4 cm
\end{itemize}

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

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

 Dans le triangle CZF rectangle en C, on sait que :
\begin{itemize}
\item CZ=8,1 cm
\item \(\widehat{ZFC}\) =  29\textdegree{}
\end{itemize}

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

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

 Dans le triangle VCD rectangle en V, on sait que :
\begin{itemize}
\item CD=4 cm
\item \(\widehat{CDV}\) =  14\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [VD]. (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 (1.448,2.129) {?};
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 Dans le triangle JHZ rectangle en J, on cherche une relation entre l'angle aigu   \(\widehat{JHZ}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{JHZ}\)) = \(\dfrac{JZ}{JH}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{JHZ}\)) = \(\dfrac{4{,}7}{1{,}4}\)
\par\addvspace{.5em}
 On a donc JHZ = ArcTan(\(\dfrac{4{,}7}{1{,}4}\)) \(\approx\) 73\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{BNR}\)) = \(\dfrac{BN}{NR}\)
\par\addvspace{.5em}
d'où cos(67\textdegree{}) = \(\dfrac{BN}{6{,}1}\)

On a donc BN = 6,1 \(\times\) cos(67\textdegree{}) \(\approx\) 2,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{ADJ}\)) = \(\dfrac{AJ}{AD}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{ADJ}\)) =  \(\dfrac{1{,}2}{5{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ADJ}\) =  ArcTan(\(\dfrac{1{,}2}{5{,}4}\)) \(\approx\) 13\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
sin(\(\widehat{CFZ}\)) = \(\dfrac{CZ}{ZF}\)
\par\addvspace{.5em}
 D'où sin(29\textdegree{}) = \(\dfrac{8{,}1}{ZF}\)
\par\addvspace{.5em}
On a donc ZF =\(\dfrac{8{,}1}{\sin(29^{\circ})}\) \(\approx\) 16,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.056) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.288,1.404) {14°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{VDC}\)) = \(\dfrac{VD}{CD}\)
\par\addvspace{.5em}
  d'où    cos(14\textdegree{}) =  \(\dfrac{VD}{4}\)

On a donc VD = 4 \(\times\) cos(14\textdegree{}) \(\approx\) 3{,}9 cm\end{document}
