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\begin{document}

\sloppy

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

 Dans le triangle LZD rectangle en L, on sait que :
\begin{itemize}
\item LZ=2,4 cm
\item \(\widehat{LZD}\) = 69\textdegree{}
\end{itemize}

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

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

 Dans le triangle BNA rectangle en B, on sait que :
\begin{itemize}
\item BA=2,7 cm
\item \(\widehat{BNA}\) = 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 VKH rectangle en V, on sait que :
\begin{itemize}
\item VH=3,9 cm
\item KH=8,2 cm
\end{itemize}

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

 Dans le triangle TFN rectangle en T, on sait que :
\begin{itemize}
\item TF=3,1 cm
\item TN=3,9 cm
\end{itemize}

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

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

 Dans le triangle BMG rectangle en B, on sait que :
\begin{itemize}
\item MG=3,1 cm
\item \(\widehat{BMG}\) = 68\textdegree{}
\end{itemize}

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

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

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

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

cos(\(\widehat{LZD}\)) = \(\dfrac{LZ}{ZD}\)
\par\addvspace{.5em}
 d'où cos(69\textdegree{}) = \(\dfrac{2{,}4}{ZD}\)
\par\addvspace{.5em}
On a donc ZD = \(\dfrac{2{,}4}{\cos(69^{\circ})}\)\(\approx\) 6,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

\par\addvspace{.5em}
tan(\(\widehat{BNA}\)) = \(\dfrac{BA}{BN}\)
\par\addvspace{.5em}
tan(67\textdegree{}) = \(\dfrac{2{,}7}{BN}\)
\par\addvspace{.5em}
 On a donc BN = \(\dfrac{2{,}7}{\tan(67^{\circ})}\) \(\approx\) 1,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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

\par\addvspace{.5em}
sin(\(\widehat{VKH}\)) =  \(\dfrac{VH}{KH}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{VKH}\)) = \(\dfrac{3{,}9}{8{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{VKH}\) =  ArcSin(\(\dfrac{3{,}9}{8{,}2}\)) \(\approx\) 28\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

\par\addvspace{.5em}
tan(\(\widehat{TNF}\)) = \(\dfrac{TF}{TN}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{TNF}\)) =  \(\dfrac{3{,}1}{3{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{TNF}\) =  ArcTan(\(\dfrac{3{,}1}{3{,}9}\)) \(\approx\) 38\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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

\par\addvspace{.5em}
sin(\(\widehat{BMG}\)) =  \(\dfrac{BG}{MG}\)
\par\addvspace{.5em}
 d'où sin(68\textdegree{}) = \(\dfrac{BG}{3{,}1}\)

On a donc BG = 3,1 \(\times\) sin(68\textdegree{}) \(\approx\) 2,9 cm\end{document}
