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\begin{document}

\sloppy

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

 Dans le triangle LZW rectangle en L, on sait que :
\begin{itemize}
\item LZ=4,3 cm
\item \(\widehat{ZWL}\) =  45\textdegree{}
\end{itemize}

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

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

 Dans le triangle GWV rectangle en G, on sait que :
\begin{itemize}
\item GV=6,4 cm
\item WV=8,6 cm
\end{itemize}

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

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

 Dans le triangle BMR rectangle en B, on sait que :
\begin{itemize}
\item MR=4,3 cm
\item \(\widehat{BMR}\) = 46\textdegree{}
\end{itemize}

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

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

 Dans le triangle NVH rectangle en N, on sait que :
\begin{itemize}
\item NV=2,2 cm
\item VH=8 cm
\end{itemize}

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

 Dans le triangle JGK rectangle en J, on sait que :
\begin{itemize}
\item JG=5,4 cm
\item \(\widehat{JGK}\) = 76\textdegree{}
\end{itemize}

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

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

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

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

\par\addvspace{.5em}
tan(\(\widehat{LWZ}\)) = \(\dfrac{LZ}{LW}\)
\par\addvspace{.5em}
d'où  tan(45\textdegree{}) =  \(\dfrac{4{,}3}{LW}\)
\par\addvspace{.5em}
On a donc LW = \(\dfrac{4{,}3}{\tan(45^{\circ})}\) \(\approx\) 4,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

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

cos(\(\widehat{GVW}\)) = \(\dfrac{GV}{WV}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{GVW}\)) =  \(\dfrac{6{,}4}{8{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{GVW}\) = Arccos (\(\dfrac{6{,}4}{8{,}6}\)) \(\approx\) 42\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{BMR}\)) = \(\dfrac{BM}{MR}\)
\par\addvspace{.5em}
d'où cos(46\textdegree{}) = \(\dfrac{BM}{4{,}3}\)

On a donc BM = 4,3 \(\times\) cos(46\textdegree{}) \(\approx\) 3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

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

cos(\(\widehat{NVH}\)) = \(\dfrac{NV}{VH}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{NVH}\)) = \(\dfrac{2{,}2}{8}\)
\par\addvspace{.5em}
On a donc \(\widehat{NVH}\) =ArcCos(\(\dfrac{2{,}2}{8}\)) \(\approx\)74\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

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\end{tikzpicture}\end{center}


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

cos(\(\widehat{JGK}\)) = \(\dfrac{JG}{GK}\)
\par\addvspace{.5em}
 d'où cos(76\textdegree{}) = \(\dfrac{5{,}4}{GK}\)
\par\addvspace{.5em}
On a donc GK = \(\dfrac{5{,}4}{\cos(76^{\circ})}\)\(\approx\) 22,3 cm\end{document}
